# Evaluate $\int \cos^3 x\;\sin^2 xdx$

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

-
Well, for one thing, you forgot the constant of integration. Otherwise, yes it looks fine. What did you get from Wolfram? It could simply be that Wolfram's answer and yours differ by a constant. – Arturo Magidin Feb 11 '11 at 21:40
integral cos^3(x) sin^2(x) dx = 1/30 sin^3(x) (3 cos(2 x)+7)+constant – Krysten Feb 11 '11 at 21:41
If you want to check if you found the anti-derivative correctly, why not try differentiating the answer you got and see if you can match those up? +1 for showing your working though. – Aryabhata Feb 11 '11 at 21:50
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. – Arturo Magidin Feb 11 '11 at 22:02
good to know. thanks – Krysten Feb 14 '11 at 20:33

\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*}

-
@night owl: First, it's not my answer, it's Sivaram's. Second, he's not integrating, he's computing a trigonometric identity, so there is no "plus $C$" to keep track of. – Arturo Magidin Mar 20 '11 at 23:57
@night owl: I am just showing that $\frac{\sin^3(x)}{3} - \frac{\sin^5(x)}{5} = \frac{\sin^3(x)}{30}(3 \cos(2x) + 7)$. I am not evaluating the integral... – user17762 Mar 20 '11 at 23:59
@Arturo: Okay I understand. – night owl Apr 22 '11 at 7:46

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*}

-

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*}

-