Evaluating $\int \sin^2 x \cos^3 x \operatorname dx$

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

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

I know there is a rule in my book (with little explanation) that tells me when I had an odd and an even degree on two trig functions I should split the odd and convert it to an identity but this way seems easier, and I can't get an answer either way.

$$\int \cos^3 dx - \int \cos^ 5 x dx$$

I am not sure where to go from here, I don't know how to get the integral of $\cos^3 x$

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The problem I am having is that I am not getting the correct answer. – user138246 Jun 1 '12 at 16:09
Whenever you have an odd number of one, say sine (but exact same thing works for cosine), and an even number of the other, say cosine, convert all but one of your sines into cosines using $\sin^2 x = 1 - \cos^2 x$. Now, you're left with a bunch of $\cos x$ terms to various degrees, and one $\sin x$. Let $u = \cos x$ and then $du = -\sin x \,dx$. You have $\sin x \,dx$ so the substitution works out easily. – Graphth Jun 1 '12 at 16:11

Hint:

• $\sin^{2}{x} \cdot \cos^{3}(x) = \sin^{2}{x} \cdot \bigl(1-\sin^{2}(x)\bigr) \cdot \cos{x}$

• Now put $t = \sin(x)$. The integral you have has the form $t^{2} \cdot (1-t^{2}) = t^{2}-t^{4}$.

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I tried that and was left with $\int sinx - sin^3 x$ which seems wrong, it complicates the problem. – user138246 Jun 1 '12 at 16:10
@Jordan: I guess you didn't follow my comment. I have made an edit. Is it helpful now. – user9413 Jun 1 '12 at 16:12

Noting that: $\int \sin^2{x}\cos^3{x}dx= \int\sin^2{x}\cos{x}dx-\int\sin^4{x}\cos{x}dx$ by the substitution $\cos^2{x}=1-\sin^2{x}$.

Then, make the substitution $u=\sin{x} \implies du=\cos{x}dx$ and the integral should be easy to calculate from there :)

(i.e. you should get $\int u^2 du - \int u^4 du$)

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There are general rules to give power of sin and cosine in terms of multiples of the angle. So, one can write $$\sin^2\theta=\frac{1}{2}(1-\cos(2\theta))$$ and $$\cos^2\theta=\frac{1}{2}(1+\cos(2\theta)).$$ Also, a formula like $\cos\alpha\cos\beta=\frac{1}{2}\left[\cos(\alpha+\beta)+\cos(\alpha-\beta)\right]$ is essential.

In your case is $$\cos^3x=\cos^2x\cos x=\frac{1}{2}(1+\cos(2x))\cos x=\frac{1}{2}(\cos x+\cos(2x)\cos x)$$ but $\cos(2x)\cos x=\frac{1}{2}(\cos(3x)+\cos(x))$ and you get $$\cos^3x=\frac{3}{4}\cos x+\frac{1}{4}\cos(2x).$$ Finally, $$\sin^2x\cos^3x=\frac{1}{2}(1-\cos(2x))\left(\frac{3}{4}\cos x+\frac{1}{4}\cos(2x)\right)=\frac{3}{8}\cos x-\frac{3}{8}\cos x\cos(2x)+\frac{1}{8}\cos(2x)-\frac{1}{8}\cos^2(2x)$$ that gives $$\sin^2x\cos^3x=-\frac{1}{16}+\frac{3}{16}\cos x+\frac{1}{8}\cos(2x)-\frac{3}{8}\cos(3x)-\frac{1}{16}\cos(4x).$$

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