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JH
Numerade Educator

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Problem 1 Easy Difficulty

Evaluate the integral.

$ \displaystyle \int \sin^2 x \cos^3 x dx $

Answer

$\frac{1}{3} \sin ^{3} x-\frac{1}{5} \sin ^{5} x+C$

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Video Transcript

This problem is from Chapter seven section to problem number one from Calculus Early Transcendental Sze eighth Edition by James Door And here we have the integral of sine square times coast and cubed of X t X. So the first thing we can do here is we observed that we have ah, odd power on the co sign. So let's first rewrite this by pulling out one of the factors of co sign. So here we are. Rewrite Cho San Cube as co sign squared times call sign. The next thing we could do is apply it. Pythagorean identity to rewrite this co sign squared is one minus sign square. Okay, so that's coming from. But dragon identity that says sine squared plus co sign squared equals one. And then now, after the stuff, we basically see that the problem is tailor made for you Substitution. So here we can apply the use up u equals sign of X so that do you becomes cosign a Vicks the ex. Okay, so then after the step, we have the integral you square one minus you squared Deal. So here we can distribute the power of you square toe one and minus. You squared before we integrate. So we have you square minus you to the fourth, do you? And now we could proceed to integration. So here we applied the power Rolls Royce. So we have you knew that their power over three minus, you know, the fifth hour over five plus c. So we have evaluated the integral. But now she's back. Substitutes are original variable x. So we simply rewrite you, Cube. But using our U substitution. So this becomes Sang Cube. And similarly, you fit becomes signs of the fifth plus e And that's where in the rule