answer：First, I need to determine the total number of students surveyed. Let's denote this total as ( T ).It's given that 60% of the students do not speak French. This means that 40% of the students do speak French. Let's denote the number of students who speak French as ( F ).So, ( F = 0.40 times T ).Among the students who speak French, 20 speak English well, and 60 do not speak English. Therefore, the total number of students who speak French is:[ F = 20 + 60 ][ F = 80 ]Now, substituting ( F ) into the equation:[ 80 = 0.40 times T ][ T = frac{80}{0.40} ][ T = 200 ]Therefore, the total number of students surveyed is 200.

answer：To determine the possible values of ( B ) that make the number ( 785,342,B52 ) divisible by 16, we focus on the last three digits, ( B52 ).1. **Divisibility Rule for 16**: A number is divisible by 16 if its last three digits form a number that is divisible by 16.2. **Form the Number**: The last three digits of ( 785,342,B52 ) are ( B52 ). We need to find the values of ( B ) (ranging from 0 to 9) such that ( B52 ) is divisible by 16.3. **Check Each Value of ( B )**: - ( 052 div 16 = 3.25 ) → Not divisible - ( 152 div 16 = 9.5 ) → Not divisible - ( 252 div 16 = 15.75 ) → Not divisible - ( 352 div 16 = 22 ) → Not divisible - ( 452 div 16 = 28.25 ) → Not divisible - ( 552 div 16 = 34.5 ) → Not divisible - ( 652 div 16 = 40.75 ) → Not divisible - ( 752 div 16 = 47 ) → Not divisible - ( 852 div 16 = 53.25 ) → Not divisible - ( 952 div 16 = 59.5 ) → Not divisible4. **Conclusion**: None of the values of ( B ) from 0 to 9 result in ( B52 ) being divisible by 16. Therefore, there are no valid values of ( B ).5. **Sum of Possible Values**: Since no valid ( B ) exists, the sum of all possible values of ( B ) is 0.

answer：Okay, so I need to figure out how many holiday shoppers the store is predicting based on the information given. Let me start by understanding the problem step by step.First, the store buys candied yams in bulk boxes, and each bulk box contains 25 packages. That’s straightforward enough. Now, it says that every third shopper buys a package of candied yams. So, out of every three shoppers, one buys a package. That means the ratio of shoppers to packages sold is 3:1.The store needs to order 5 bulk boxes. Since each bulk box has 25 packages, I can calculate the total number of packages by multiplying the number of bulk boxes by the number of packages per box. So, 5 bulk boxes times 25 packages per box equals 125 packages. That means the store needs to have 125 packages of candied yams to meet the demand.Now, since every third shopper buys a package, I need to find out how many shoppers would result in 125 packages being sold. If one out of every three shoppers buys a package, then the total number of shoppers would be three times the number of packages sold. So, I take the total number of packages, which is 125, and multiply it by 3 to get the total number of shoppers.Let me write that out: 125 packages * 3 shoppers per package = 375 shoppers. So, the store is predicting that 375 holiday shoppers will come in, and out of those, every third one will buy a package of candied yams, resulting in the need for 125 packages, which requires 5 bulk boxes.Wait, let me double-check that to make sure I didn't make a mistake. If there are 375 shoppers and every third shopper buys a package, then 375 divided by 3 should give me the number of packages sold. 375 / 3 = 125. Yes, that matches up with the number of packages needed. So, it seems like my calculation is correct.I think I've got it. The store needs to order 5 bulk boxes, which gives them 125 packages, and since every third shopper buys a package, they're expecting 375 holiday shoppers.

answer：Alright, so I have this geometry problem here. It involves a triangle ABC with two altitudes, AD and BE, intersecting at point H. I'm given that angle BAC is 58 degrees and angle ABC is 67 degrees. I need to find the measure of angle AHB.First, I should probably sketch a rough diagram to visualize the problem. Triangle ABC, with points A, B, and C. AD is an altitude from A to BC, and BE is an altitude from B to AC. These two altitudes intersect at point H, which is the orthocenter of the triangle.Okay, so I know that in any triangle, the sum of the interior angles is 180 degrees. Since I'm given two angles, BAC and ABC, I can find the third angle, which is angle ACB. Let me calculate that.Angle BAC is 58 degrees, angle ABC is 67 degrees. So angle ACB should be 180 - 58 - 67. Let me do the math: 58 + 67 is 125, so 180 - 125 is 55 degrees. So angle ACB is 55 degrees.Now, I need to find angle AHB. Hmm, angle AHB is the angle at point H between points A and B. Since H is the orthocenter, it's where the altitudes meet. I remember that in a triangle, the orthocenter has some interesting properties, especially related to the angles of the triangle.I think there's a relationship between the angles of the triangle and the angles formed at the orthocenter. Maybe something about cyclic quadrilaterals? Let me recall. If two altitudes intersect, they form right angles with the sides of the triangle. So, AD is perpendicular to BC, and BE is perpendicular to AC.Wait, so quadrilateral ADEH is formed by the feet of the altitudes and the orthocenter. Since both AD and BE are altitudes, they are perpendicular to the opposite sides. So, angles ADE and AEB are right angles. That makes quadrilateral ADEH a cyclic quadrilateral because it has two opposite right angles. In a cyclic quadrilateral, the sum of opposite angles is 180 degrees.So, if quadrilateral ADEH is cyclic, then angle AHB is supplementary to angle ACB. That is, angle AHB + angle ACB = 180 degrees. Since I already found angle ACB to be 55 degrees, angle AHB should be 180 - 55, which is 125 degrees.Wait, let me make sure I'm not mixing up any properties here. Is angle AHB really supplementary to angle ACB? Let me think about the cyclic quadrilateral again. In cyclic quadrilateral ADEH, points A, D, E, and H lie on a circle. So, angle AHB is the angle subtended by arc AB at point H, and angle ACB is the angle subtended by the same arc AB at point C.But in a circle, angles subtended by the same arc are equal if they are on the same side of the arc. However, since H is inside the triangle and C is outside the circle (I think), maybe they are supplementary instead. Hmm, I need to clarify this.Alternatively, maybe I can use the fact that in triangle ABC, the orthocenter H creates several smaller triangles, each of which has some known angle relationships. For example, in triangle AHB, the angles at A and B are related to the original angles of triangle ABC.Wait, another approach: since AD and BE are altitudes, they are perpendicular to BC and AC respectively. So, angles ADB and BEC are right angles. Therefore, quadrilateral ADEH has two right angles at D and E. Hence, it's cyclic because the sum of opposite angles in a cyclic quadrilateral is 180 degrees, and here we have two right angles which add up to 180 degrees.So, in cyclic quadrilateral ADEH, angle AHB is equal to 180 degrees minus angle ACB. Since angle ACB is 55 degrees, angle AHB is 180 - 55 = 125 degrees.I think that makes sense. Let me double-check. If I consider the cyclic quadrilateral ADEH, then angle AHB and angle ACB are supplementary because they are opposite angles in the cyclic quadrilateral. Therefore, angle AHB is 125 degrees.Yeah, that seems consistent. I don't think I made any mistakes in the reasoning. So, I'm pretty confident that angle AHB is 125 degrees.**Final Answer**boxed{125^circ}

answer：First, I need to determine how many votes Geoff received. Since he got 1% of the 6,000 votes cast, I calculate 1% of 6,000.Next, I know that Geoff needed exactly 3,000 more votes to win the election. By adding these 3,000 votes to the votes he already received, I can find the total number of votes required to win.Finally, to find the required percentage to win the election, I divide the total votes needed by the total votes cast and then multiply by 100 to convert it to a percentage.

answer：Okay, so I have this integral to solve: ∫ dx / [sin(x - 1) sin(x - 2)]. The hint suggests using the substitution tan x = t. Hmm, I'm not entirely sure how that substitution will help, but let me try to work through it step by step.First, I remember that when dealing with integrals involving trigonometric functions, sometimes using trigonometric identities can simplify the expression. The hint mentions using tan x = t, which is a substitution that often helps in integrals involving sine and cosine. Maybe I can express everything in terms of tan x and see if that simplifies the integral.Let me write down the integral again:∫ dx / [sin(x - 1) sin(x - 2)]I notice that both denominators are sine functions with arguments shifted by constants. Maybe I can use some identity to combine these terms or simplify the product of sines.I recall that the product of sines can be expressed using the cosine of the sum and difference of angles. Specifically, the identity:sin A sin B = [cos(A - B) - cos(A + B)] / 2Let me apply this identity to the denominator:sin(x - 1) sin(x - 2) = [cos((x - 1) - (x - 2)) - cos((x - 1) + (x - 2))] / 2Simplifying the arguments inside the cosines:(x - 1) - (x - 2) = x - 1 - x + 2 = 1(x - 1) + (x - 2) = 2x - 3So, the denominator becomes:[cos(1) - cos(2x - 3)] / 2Therefore, the integral becomes:∫ dx / [ (cos(1) - cos(2x - 3)) / 2 ] = 2 ∫ dx / [cos(1) - cos(2x - 3)]Hmm, that seems a bit simpler, but I still have an integral involving cos(2x - 3). Maybe I can use another substitution here. Let me let u = 2x - 3. Then, du/dx = 2, so dx = du/2.Substituting, the integral becomes:2 ∫ (du/2) / [cos(1) - cos(u)] = ∫ du / [cos(1) - cos(u)]Now, I have to evaluate ∫ du / [cos(1) - cos(u)]. This looks like a standard integral that might have a known form. I think there's an identity or a substitution that can help here.I remember that integrals of the form ∫ du / [a - cos(u)] can be evaluated using the Weierstrass substitution, which is t = tan(u/2). Let me try that.Let t = tan(u/2). Then, we have:sin(u) = 2t / (1 + t²)cos(u) = (1 - t²) / (1 + t²)du = 2 dt / (1 + t²)Substituting these into the integral:∫ [2 dt / (1 + t²)] / [cos(1) - (1 - t²)/(1 + t²)]Simplify the denominator:cos(1) - (1 - t²)/(1 + t²) = [cos(1)(1 + t²) - (1 - t²)] / (1 + t²)= [cos(1) + cos(1) t² - 1 + t²] / (1 + t²)= [ (cos(1) - 1) + (cos(1) + 1) t² ] / (1 + t²)So, the integral becomes:∫ [2 dt / (1 + t²)] / [ (cos(1) - 1) + (cos(1) + 1) t² ) / (1 + t²) ]The (1 + t²) terms cancel out:∫ 2 dt / [ (cos(1) - 1) + (cos(1) + 1) t² ]Let me factor out the constants:= 2 ∫ dt / [ (cos(1) - 1) + (cos(1) + 1) t² ]Let me denote A = cos(1) - 1 and B = cos(1) + 1 for simplicity:= 2 ∫ dt / (A + B t²)This is a standard integral of the form ∫ dt / (a + b t²) = (1/√(ab)) arctan(t √(b/a)) + CSo, applying this formula:= 2 * [1 / √(A B)] arctan(t √(B/A)) + CNow, substituting back A and B:A = cos(1) - 1B = cos(1) + 1So, √(A B) = √[(cos(1) - 1)(cos(1) + 1)] = √[cos²(1) - 1] = √[-sin²(1)] = i sin(1)Wait, that's imaginary, which doesn't make sense in this context. I must have made a mistake somewhere.Let me check my steps. When I used the Weierstrass substitution, I ended up with an integral that led to an imaginary number, which shouldn't happen because the original integral is real. Maybe I made a mistake in simplifying the denominator.Going back to the substitution step:After substituting t = tan(u/2), I had:∫ [2 dt / (1 + t²)] / [cos(1) - (1 - t²)/(1 + t²)]Let me re-express the denominator:cos(1) - (1 - t²)/(1 + t²) = [cos(1)(1 + t²) - (1 - t²)] / (1 + t²)= [cos(1) + cos(1) t² - 1 + t²] / (1 + t²)= [ (cos(1) - 1) + (cos(1) + 1) t² ] / (1 + t²)Wait, cos(1) - 1 is negative because cos(1) is less than 1. So, A = cos(1) - 1 is negative, and B = cos(1) + 1 is positive. Therefore, A * B is negative, which means √(A B) is imaginary. That's the issue.Hmm, maybe I should approach this integral differently. Instead of using the Weierstrass substitution, perhaps I can use a different identity or substitution.Let me think. The integral is ∫ du / [cos(1) - cos(u)]. I remember that there's an identity for the difference of cosines:cos A - cos B = -2 sin[(A + B)/2] sin[(A - B)/2]So, applying this to the denominator:cos(1) - cos(u) = -2 sin[(1 + u)/2] sin[(1 - u)/2]Therefore, the integral becomes:∫ du / [ -2 sin((1 + u)/2) sin((1 - u)/2) ]= -1/2 ∫ du / [ sin((1 + u)/2) sin((1 - u)/2) ]Hmm, that seems more complicated. Maybe I can use another substitution here.Let me try to express the denominator in terms of sine functions. Alternatively, perhaps I can use the identity for the product of sines again.Wait, I just used that identity to get here. Maybe I need a different approach.Let me consider another substitution. Let me set v = u - 1. Then, u = v + 1, and du = dv.Substituting into the integral:∫ dv / [cos(1) - cos(v + 1)]Now, using the identity cos(v + 1) = cos v cos 1 - sin v sin 1.So, the denominator becomes:cos(1) - [cos v cos 1 - sin v sin 1] = cos(1) - cos v cos 1 + sin v sin 1= cos(1)(1 - cos v) + sin v sin 1Hmm, not sure if that helps. Maybe I can factor something out.Alternatively, perhaps I can use the identity for the difference of cosines again.Wait, let's try to express the denominator as:cos(1) - cos(u) = 2 sin[(u + 1)/2] sin[(u - 1)/2]Wait, earlier I had cos A - cos B = -2 sin[(A + B)/2] sin[(A - B)/2], so cos(1) - cos(u) = -2 sin[(1 + u)/2] sin[(1 - u)/2]So, the integral becomes:∫ du / [ -2 sin((1 + u)/2) sin((1 - u)/2) ]= -1/2 ∫ du / [ sin((1 + u)/2) sin((1 - u)/2) ]Let me make a substitution here. Let me set t = (u - 1)/2. Then, u = 2t + 1, and du = 2 dt.Substituting into the integral:-1/2 ∫ 2 dt / [ sin((1 + 2t + 1)/2) sin((1 - (2t + 1))/2) ]Simplify the arguments:(1 + 2t + 1)/2 = (2 + 2t)/2 = 1 + t(1 - (2t + 1))/2 = (1 - 2t - 1)/2 = (-2t)/2 = -tSo, the integral becomes:-1/2 * 2 ∫ dt / [ sin(1 + t) sin(-t) ]Simplify:-1/2 * 2 = -1, and sin(-t) = -sin t, so:-1 ∫ dt / [ sin(1 + t) (-sin t) ] = ∫ dt / [ sin(1 + t) sin t ]Hmm, now I have ∫ dt / [ sin t sin(1 + t) ]. This looks similar to the original integral but with shifted arguments. Maybe I can use the same approach as before.Let me apply the identity sin A sin B = [cos(A - B) - cos(A + B)] / 2 again.So, sin t sin(1 + t) = [cos(t - (1 + t)) - cos(t + (1 + t))] / 2Simplify the arguments:t - (1 + t) = -1t + (1 + t) = 2t + 1So, sin t sin(1 + t) = [cos(-1) - cos(2t + 1)] / 2 = [cos(1) - cos(2t + 1)] / 2Therefore, the integral becomes:∫ dt / [ (cos(1) - cos(2t + 1)) / 2 ] = 2 ∫ dt / [cos(1) - cos(2t + 1)]This seems like we're going in circles. Maybe I need a different strategy.Wait, perhaps instead of using the Weierstrass substitution, I can use a different substitution. Let me try to express the integral in terms of cotangent or something else.Let me recall that ∫ dx / sin x = ln |tan(x/2)| + C. Maybe I can manipulate the integral to express it in terms of cotangent or tangent.Alternatively, perhaps I can use partial fractions. Let me see if I can express 1 / [sin(x - 1) sin(x - 2)] as a combination of simpler fractions.I remember that 1 / [sin A sin B] can sometimes be expressed as a combination of cot A and cot B. Let me try that.Let me assume that:1 / [sin(x - 1) sin(x - 2)] = A cot(x - 1) + B cot(x - 2)Let me solve for A and B.Multiplying both sides by sin(x - 1) sin(x - 2):1 = A sin(x - 2) + B sin(x - 1)Hmm, this seems tricky because it's supposed to hold for all x, but the right side is a combination of sine functions, which are periodic and not constants. Therefore, this approach might not work.Alternatively, maybe I can use the identity for the difference of angles. Let me consider:sin(x - 1) - sin(x - 2) = 2 cos[(x - 1 + x - 2)/2] sin[(x - 1 - (x - 2))/2] = 2 cos(x - 1.5) sin(0.5)So, sin(x - 1) - sin(x - 2) = 2 sin(0.5) cos(x - 1.5)Therefore, 1 = [sin(x - 1) - sin(x - 2)] / [2 sin(0.5) cos(x - 1.5)]Hmm, not sure if that helps directly, but maybe I can use this to express 1 / [sin(x - 1) sin(x - 2)] in terms of cotangent.Wait, let me think differently. Let me consider the original integral:∫ dx / [sin(x - 1) sin(x - 2)]Let me make a substitution to simplify the arguments. Let me set y = x - 1.5, so that x = y + 1.5. Then, x - 1 = y + 0.5 and x - 2 = y - 0.5.So, the integral becomes:∫ dy / [sin(y + 0.5) sin(y - 0.5)]Now, using the identity sin(A + B) sin(A - B) = sin² A - sin² BWait, actually, sin(A + B) sin(A - B) = sin² A - sin² BSo, sin(y + 0.5) sin(y - 0.5) = sin² y - sin² 0.5Therefore, the integral becomes:∫ dy / [sin² y - sin² 0.5]This looks like a standard integral. I think the integral of 1 / (sin² y - a²) dy can be expressed in terms of logarithmic functions.Let me recall that ∫ dy / (sin² y - a²) can be handled using substitution or partial fractions.Let me write it as:∫ dy / [sin² y - sin² 0.5] = ∫ dy / [sin² y - k²], where k = sin 0.5I think this can be expressed using the identity:1 / (sin² y - k²) = [1 / (2k)] [1 / (sin y - k) - 1 / (sin y + k)]Wait, let me check:Let me assume that 1 / (sin² y - k²) = A / (sin y - k) + B / (sin y + k)Multiplying both sides by (sin² y - k²):1 = A (sin y + k) + B (sin y - k)= (A + B) sin y + (A k - B k)For this to hold for all y, the coefficients of sin y and the constant term must be zero except for the constant term on the left side.So, we have:A + B = 0A k - B k = 1From the first equation, B = -ASubstituting into the second equation:A k - (-A) k = 1 => A k + A k = 1 => 2 A k = 1 => A = 1 / (2k)Therefore, B = -1 / (2k)So, 1 / (sin² y - k²) = [1 / (2k)] [1 / (sin y - k) - 1 / (sin y + k)]Therefore, the integral becomes:∫ [1 / (2k)] [1 / (sin y - k) - 1 / (sin y + k)] dy= 1/(2k) [ ∫ dy / (sin y - k) - ∫ dy / (sin y + k) ]Now, I need to evaluate ∫ dy / (sin y - k) and ∫ dy / (sin y + k)I recall that integrals of the form ∫ dy / (sin y ± a) can be evaluated using the substitution t = tan(y/2), which is the Weierstrass substitution.Let me try that for ∫ dy / (sin y - k)Let t = tan(y/2). Then, sin y = 2t / (1 + t²), and dy = 2 dt / (1 + t²)Substituting into the integral:∫ [2 dt / (1 + t²)] / [2t / (1 + t²) - k]= ∫ 2 dt / [2t - k(1 + t²)]= ∫ 2 dt / [ -k t² + 2t - k ]Let me write the denominator as -k t² + 2t - k = - (k t² - 2t + k)So, the integral becomes:∫ 2 dt / [ - (k t² - 2t + k) ] = -2 ∫ dt / (k t² - 2t + k)This is a quadratic in the denominator. Let me complete the square:k t² - 2t + k = k (t² - (2/k) t) + k= k [ t² - (2/k) t + (1/k²) ] - k (1/k²) + k= k [ (t - 1/k)² ] - 1/k + k= k (t - 1/k)² + (k - 1/k)Hmm, this seems a bit messy. Maybe I can factor the quadratic.Alternatively, let me consider the integral:∫ dt / (a t² + b t + c)The standard form is ∫ dt / (a t² + b t + c) = [1 / √(4ac - b²)] ln | (2a t + b) / √(a) + √(4ac - b²) | + C, if 4ac - b² > 0In our case, a = k, b = -2, c = kSo, 4ac - b² = 4k * k - (-2)^2 = 4k² - 4 = 4(k² - 1)Since k = sin 0.5, and sin 0.5 is less than 1, so k² - 1 is negative. Therefore, 4ac - b² is negative, which means we have to use a different form involving arctangent.Wait, no, actually, if 4ac - b² is negative, the integral involves logarithms with complex arguments, which complicates things. Maybe I made a mistake in the substitution.Alternatively, perhaps I can use a different substitution. Let me try to express the denominator as a perfect square.Wait, let me go back to the integral:∫ dy / (sin y - k)Let me use the substitution t = tan(y/2), which gives sin y = 2t / (1 + t²), dy = 2 dt / (1 + t²)So, the integral becomes:∫ [2 dt / (1 + t²)] / [2t / (1 + t²) - k]= ∫ 2 dt / [2t - k(1 + t²)]= ∫ 2 dt / [ -k t² + 2t - k ]Let me factor out -k from the denominator:= ∫ 2 dt / [ -k (t² - (2/k) t + 1) ]= -2/k ∫ dt / (t² - (2/k) t + 1)Now, complete the square in the denominator:t² - (2/k) t + 1 = (t - 1/k)^2 + 1 - 1/k²= (t - 1/k)^2 + (k² - 1)/k²So, the integral becomes:-2/k ∫ dt / [ (t - 1/k)^2 + (k² - 1)/k² ]Let me set u = t - 1/k, then du = dtThe integral becomes:-2/k ∫ du / [ u² + (k² - 1)/k² ]= -2/k ∫ du / [ u² + ( (k² - 1)/k² ) ]This is of the form ∫ du / (u² + a²) = (1/a) arctan(u/a) + CSo, here, a² = (k² - 1)/k², so a = sqrt( (k² - 1)/k² ) = sqrt(k² - 1)/kBut since k = sin 0.5, and sin 0.5 < 1, k² - 1 is negative, so a is imaginary. Hmm, this is getting complicated.Wait, maybe I should approach this differently. Let me consider that since k = sin 0.5, which is a constant, perhaps I can express the integral in terms of logarithmic functions without going through complex numbers.Alternatively, perhaps I can use the substitution z = tan(y/2 + α), where α is chosen such that the integral simplifies. This is a method used for integrals involving sin y and constants.Let me try that. Let me set z = tan(y/2 + α), where α is chosen to simplify the denominator.But this might be too involved. Maybe I should look for another approach.Wait, going back to the integral:∫ dy / [sin² y - k²] = ∫ dy / [sin² y - sin² 0.5]I remember that sin² y - sin² a = sin(y - a) sin(y + a)So, sin² y - sin² 0.5 = sin(y - 0.5) sin(y + 0.5)Therefore, the integral becomes:∫ dy / [sin(y - 0.5) sin(y + 0.5)]Wait, this is similar to the original integral but with shifted arguments. Maybe I can use the same approach as before.Let me apply the identity sin A sin B = [cos(A - B) - cos(A + B)] / 2So, sin(y - 0.5) sin(y + 0.5) = [cos((y - 0.5) - (y + 0.5)) - cos((y - 0.5) + (y + 0.5))] / 2Simplify the arguments:(y - 0.5) - (y + 0.5) = -1(y - 0.5) + (y + 0.5) = 2ySo, sin(y - 0.5) sin(y + 0.5) = [cos(-1) - cos(2y)] / 2 = [cos(1) - cos(2y)] / 2Therefore, the integral becomes:∫ dy / [ (cos(1) - cos(2y)) / 2 ] = 2 ∫ dy / [cos(1) - cos(2y)]This is similar to the integral I had earlier. Let me make a substitution here. Let me set u = 2y, so y = u/2, dy = du/2Substituting, the integral becomes:2 ∫ (du/2) / [cos(1) - cos(u)] = ∫ du / [cos(1) - cos(u)]Wait, this is the same integral I had before. So, it seems like I'm going in circles. Maybe I need to accept that this integral will involve logarithmic functions and proceed accordingly.Let me recall that ∫ du / [cos(1) - cos(u)] can be expressed in terms of logarithmic functions. Let me try to find a substitution that can help me express this integral in terms of logarithms.Let me consider the substitution t = tan(u/2). Then, as before, sin u = 2t / (1 + t²), cos u = (1 - t²)/(1 + t²), and du = 2 dt / (1 + t²)Substituting into the integral:∫ [2 dt / (1 + t²)] / [cos(1) - (1 - t²)/(1 + t²)]= ∫ [2 dt / (1 + t²)] / [ (cos(1)(1 + t²) - (1 - t²)) / (1 + t²) ]= ∫ 2 dt / [cos(1)(1 + t²) - (1 - t²)]Simplify the denominator:cos(1)(1 + t²) - (1 - t²) = cos(1) + cos(1) t² - 1 + t²= (cos(1) - 1) + (cos(1) + 1) t²So, the integral becomes:∫ 2 dt / [ (cos(1) - 1) + (cos(1) + 1) t² ]Let me factor out the constants:= 2 ∫ dt / [ (cos(1) - 1) + (cos(1) + 1) t² ]Let me denote A = cos(1) - 1 and B = cos(1) + 1So, the integral becomes:2 ∫ dt / (A + B t²)This is a standard integral. The integral of 1 / (A + B t²) dt is (1 / √(AB)) arctan(t √(B/A)) + C, provided that A and B are positive. However, in our case, A = cos(1) - 1 is negative because cos(1) < 1, and B = cos(1) + 1 is positive. Therefore, √(AB) is imaginary, which complicates things.Hmm, this suggests that the integral might involve logarithmic functions instead of arctangent. Let me consider expressing the denominator as a difference of squares.Wait, since A is negative, let me write A = -|A|, so:2 ∫ dt / (-|A| + B t²) = 2 ∫ dt / (B t² - |A| )= 2 ∫ dt / (B t² - |A| )This is of the form ∫ dt / (a t² - b), which can be expressed using logarithms:∫ dt / (a t² - b) = (1 / (2√(ab))) ln | (√a t - √b) / (√a t + √b) | + CSo, applying this formula:Let a = B, b = |A|Then, √(ab) = √(B |A|) = √[ (cos(1) + 1)(1 - cos(1)) ] = √[1 - cos²(1)] = sin(1)Therefore, the integral becomes:2 * [1 / (2 sin(1))] ln | (√B t - √|A|) / (√B t + √|A|) | + CSimplify:= (1 / sin(1)) ln | (sqrt(B) t - sqrt(|A|)) / (sqrt(B) t + sqrt(|A|)) | + CNow, substituting back A and B:sqrt(B) = sqrt(cos(1) + 1)sqrt(|A|) = sqrt(1 - cos(1))So, the expression inside the logarithm is:( sqrt(cos(1) + 1) t - sqrt(1 - cos(1)) ) / ( sqrt(cos(1) + 1) t + sqrt(1 - cos(1)) )Let me simplify this expression. Notice that:sqrt(cos(1) + 1) = sqrt(2 cos²(0.5)) = sqrt(2) cos(0.5)Similarly, sqrt(1 - cos(1)) = sqrt(2 sin²(0.5)) = sqrt(2) sin(0.5)Therefore, the expression becomes:( sqrt(2) cos(0.5) t - sqrt(2) sin(0.5) ) / ( sqrt(2) cos(0.5) t + sqrt(2) sin(0.5) )Factor out sqrt(2):= sqrt(2) [ cos(0.5) t - sin(0.5) ] / sqrt(2) [ cos(0.5) t + sin(0.5) ]= [ cos(0.5) t - sin(0.5) ] / [ cos(0.5) t + sin(0.5) ]So, the integral becomes:(1 / sin(1)) ln | [ cos(0.5) t - sin(0.5) ] / [ cos(0.5) t + sin(0.5) ] | + CNow, recall that t = tan(u/2), and u = 2y, and y = x - 1.5So, t = tan(u/2) = tan(y) = tan(x - 1.5)Therefore, substituting back:= (1 / sin(1)) ln | [ cos(0.5) tan(x - 1.5) - sin(0.5) ] / [ cos(0.5) tan(x - 1.5) + sin(0.5) ] | + CThis expression can be simplified further. Let me factor out cos(0.5) from numerator and denominator:= (1 / sin(1)) ln | [ cos(0.5) (tan(x - 1.5) - tan(0.5)) ] / [ cos(0.5) (tan(x - 1.5) + tan(0.5)) ] | + CThe cos(0.5) terms cancel out:= (1 / sin(1)) ln | (tan(x - 1.5) - tan(0.5)) / (tan(x - 1.5) + tan(0.5)) | + CNow, I can use the identity for the difference of tangents:tan A - tan B = sin(A - B) / (cos A cos B)Similarly, tan A + tan B = sin(A + B) / (cos A cos B)But perhaps a better approach is to express the ratio in terms of sine and cosine.Let me write tan(x - 1.5) as sin(x - 1.5)/cos(x - 1.5), and tan(0.5) as sin(0.5)/cos(0.5)So, the expression inside the logarithm becomes:[ sin(x - 1.5)/cos(x - 1.5) - sin(0.5)/cos(0.5) ] / [ sin(x - 1.5)/cos(x - 1.5) + sin(0.5)/cos(0.5) ]To combine the terms, let me find a common denominator:Numerator:[ sin(x - 1.5) cos(0.5) - sin(0.5) cos(x - 1.5) ] / [ cos(x - 1.5) cos(0.5) ]Denominator:[ sin(x - 1.5) cos(0.5) + sin(0.5) cos(x - 1.5) ] / [ cos(x - 1.5) cos(0.5) ]So, the ratio becomes:[ sin(x - 1.5) cos(0.5) - sin(0.5) cos(x - 1.5) ] / [ sin(x - 1.5) cos(0.5) + sin(0.5) cos(x - 1.5) ]Notice that the numerator is sin(x - 1.5 - 0.5) = sin(x - 2)And the denominator is sin(x - 1.5 + 0.5) = sin(x - 1)Therefore, the expression simplifies to:sin(x - 2) / sin(x - 1)So, the integral becomes:(1 / sin(1)) ln | sin(x - 2) / sin(x - 1) | + CTherefore, the final answer is:(1 / sin(1)) ln | sin(x - 2) / sin(x - 1) | + CI think this makes sense. Let me check the steps to ensure I didn't make any mistakes. Starting from the substitution, using trigonometric identities, and simplifying step by step, it seems consistent. The key was recognizing that the integral could be expressed in terms of logarithmic functions after a series of substitutions and simplifications.