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Is this correct? I thought it would be but when I entered it into wolfram alpha, I got a different answer.

$\int (\cos^3x)(\sin^2x)dx = \int(\cos x)(\cos^2x)(\sin^2x)dx = \int (\cos x)(1-\sin^2x)(\sin^2x)dx.$

let $u = \sin x$, $du = \cos xdx$

$\int(1-u^2)u^2du = \int(u^2-u^4)du = \frac{u^3}{3} - \frac{u^5}{5} +C$

Plugging in back $u$, we get $\displaystyle\frac{\sin^3 x}{3} - \frac{\sin^5 x}{5}$ + C

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    I've added the $+C$ to the integral of $u$ as well; it is a common tactic for some students to just add "$+C$" at the end of all the calculations; in fact, it should go in *as soon* as you are done with the integration. When you get to differential equations, if you wait to put in the constant of integration until the end, instead of when you should, you'll get incorrect answers.2011-02-11

3 Answers 3

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\begin{align*} \frac{\sin^3 (x)}{3} - \frac{\sin^5 (x)}{5} &= \sin^3 (x) (\frac{1}{3} - \frac{\sin^2 (x)}{5});\\ \cos(2x) &= 1 - 2 \sin^2(x)\\ \sin^2(x) &= \frac{1- \cos(2x)}{2}. \end{align*}

Hence, we get,

\begin{align*} \sin^3 (x) \left(\frac{1}{3} - \frac{\sin^2 (x)}{5}\right) &= \sin^3 (x) \left(\frac{1}{3} - \frac{1 - \cos(2x)}{10}\right)\\ & = \sin^3(x) \left(\frac{10 - 3 + 3 \cos(2x)}{30}\right)\\ & = \frac{\sin^3(x)}{30} (3 \cos(2x)+7) \end{align*}

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    @Arturo: Okay I understand.2011-04-22
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Just for the heck of it, another substitution could have been $u=\sin^3 x,$ in which case, $du = 3\sin^2x\cos x \ dx.$ Now, $1-u^{2/3} = 1-\sin^2x = \cos^2x.$ Thus, $ \begin{align*} \int \cos^3x\sin^2x \ dx &= \frac{1}{3}\int \cos^2x(3\sin^2x\cos x) dx \\ &= \frac{1}{3} \int(1-u^{2/3})du \\ &= \frac{u}{3} - \frac{u^{5/3}}{5} + C \\ &= \frac{\sin^3x}{3} - \frac{\sin^5x}{5} + C \ . \end{align*} $

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Take $\sin x=t, \cos x dx=dt$ \begin{align*} \therefore \int \cos^{3}x\sin^{2}x dx &=\int \cos^{2}x\sin^{2} x\cos x dx\ &=\int (1-\sin^{2}x) \sin^{2} x \cos x dx\ &= \int (1-t^{2})t^{2} dt\ &=\int (t^{2}-t^{4})dt\ &=\frac{t^{3}}{3}-\frac{t^{5}}{5}+c\ &= \frac{\sin^{3}x}{3}-\frac{\sin^{5}x}{5}+c \end{align*}