MATH 181 Study Guide - Final Guide: Hypotenuse, Improper Integral, Numerical Integration
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Solution: we begin by rewriting sin3(x) as sin3(x) = sin(x) sin2(x) = sin(x)(1 cos2(x)). Then, after using the substitution we have the following result: u = cos(x), du = sin(x) dx. Z sin3(x)pcos(x) dx = z sin(x)(1 cos2(x))pcos(x) dx, = z (cid:16)u1/2 u5/2(cid:17) du, (cos(x))3/2 + u7/2 + c, u3/2 + 7 (cos(x))7/2 + c: integrals, part ii (trigonometric substitutions: 6 points). Do not forget to simplify your answer completely the nal answer should not contain inverse trigonometric functions! Using the above substitution, the integral transforms as follows: x = 2 tan , dx = 2 sec2 d . 1 (4 + x2)3/2 (4 + (2 tan )2)3/2 2 sec2 d . Finally, we write the answer in terms of x. To do this, we return to the substitution x = 2 tan and rewrite the equation as tan = x.
