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Math 242: Problem 25 of Sec. 8.3 8.3.25. Evaluate the integral Z 8 dx (4x2 + 1)2 Solution: We started this problem correctly in class, but I made a mistake during the computation, so here is the full solution (with many more details than you need to write, once you are comfortable with these problems). The integral is Z 8 dx = (4x2 + 1)2 Z Z 8 dx 4 x2 + 1 4 2 = 8 dx 1 = 1 2 2 2 16 x + 4 Z dx x2 + . 1 2 4 The substitution rule on the top of page 453 tells us how to deal with this kind of integral. We identify a2 = 1/4, or a = 1/2, so 1 tan θ. Then dx = 2 !2 2 1 1 1 2 1 tan θ + = tan2 θ + = 2 4 4 4 Let x = 1 2 x2 + = 4 1 sec2 θ dθ, and 2 1 1 1 (tan2 θ+1)2 = (sec2 θ)2 = sec4 θ. 16 16 16 So the integral becomes Z Z Z Z Z dx (1/2) sec2 θ dθ 16 sec2 θ dθ dθ 1 1 = =4 = 4 cos2 θ dθ. 2 = 2 2 (1/16) sec4 θ 2·2 sec4 θ sec2 θ x2 + 1 4 To evaluate R cos2 θ dθ, use the trigonometric identity Z Z 1 + cos 2θ 1 + cos 2θ 2 2 cos θ = to write 4 cos θ dθ = 4 dθ, 2 2 which is equal to Z 2 1 + cos 2θ dθ = 2θ + sin 2θ + C = 2θ + 2 sin θ cos θ + C. To get the last equality, recall the two formulas I suggested you memorize: sin(A + B) = sin A cos B + sin B cos A and cos(A + B) = cos A cos B − sin A sin B. If we let A = B = θ in the first of these, we get the “double angle” formula sin 2θ = 2 sin θ cos θ. continued on next page... 1 Math 242: Problem 25 of Sec. 8.3 To finish the solution, we must write 2θ + 2 sin θ cos θ + C in terms of x, using our substitution, x = 12 tan θ. First, to replace θ, we use x= 1 tan θ 2 ⇔ 2x = tan θ ⇔ tan−1 (2x) = θ. So 2θ = 2 tan−1 (2x). Finally, to write 2 sin θ cos θ in terms of x, draw a triangle. Since tangent is opposite over adjacent, and tan θ = 2x = 2x 1 , label the opposite edge 2x and label the adjacent edge 1. Then the √ hypotenuse must be 4x2 + 1, by the Pythagorean theorem. Since sine is opposite over hypotenuse, sin θ = √4x2x2 +1 . Cosine is adjacent over hypotenuse, so cos θ = √4x12 +1 . Putting it all together, 2x 1 4x √ 2θ + 2 sin θ cos θ + C = 2 tan−1 (2x) + 2 √ + C. + C = 2 tan−1 (2x) + 2 2 2 4x + 1 4x + 1 4x + 1 Therefore, the answer is Z 8 dx 4x = 2 tan−1 (2x) + 2 + C. (4x2 + 1)2 4x + 1 Incidentally, you can check this answer using Maxima. Use the Calculus menu, select Integrate and enter the expression 8/(4*x^2 + 1)^2 (don’t forget the *!). Maxima says the answer is atan (2 x) x 8 + , 4 8 x2 + 2 which, except for the constant C, is the same as our answer. 2