
Yefim S. answered 03/17/24
Math Tutor with Experience
sinx = u; x = π/6, u = 1/2; x = π/2, u = 1; cosxdx = du
∫π/6π/2cos5x/sin6xdx = ∫1/21(1 - u2)2/u6du =∫1/21(u-6 - 2u-4 + u-2)du = (u-5/(-5) - 2u-3/(-3) + u-1/(-1))1/21 = - 8/15 + 46/15 = 38/15
Yefim S. answered 03/17/24
Math Tutor with Experience
sinx = u; x = π/6, u = 1/2; x = π/2, u = 1; cosxdx = du
∫π/6π/2cos5x/sin6xdx = ∫1/21(1 - u2)2/u6du =∫1/21(u-6 - 2u-4 + u-2)du = (u-5/(-5) - 2u-3/(-3) + u-1/(-1))1/21 = - 8/15 + 46/15 = 38/15
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