Integrating a trigonometric polynomial by changing variable

Question

Let the integral be:

$$\int{\sin^4(x)\cos^3(x) dx} $$

I have to integrate this function by changing the variable. I'm trying: $u=\sin(x) $ and so $du = \cos(x)dx$. By rewriting the integral I get:

$$ \int{u^4 \cos^2(x)du} $$

But I'm stuck here because I'm not sure there should be any expression with x left in the integral.

Also I know the final answer is :

$$ \int{\sin^4(x)\cos^3(x)dx} = -\frac{\sin^7(x)}{7} +\frac{\sin^5(x)}{5} $$

Answer 1

Hint:

Use the identity $$\cos^{2}(x)=1-\sin^2(x)$$

and don't forget the constant of integration!

Answer 2

As the exponent of $\cos$ is odd, you can set $$u=\sin x,\quad\mathrm d u=\cos x\,\mathrm dx.$$ If the function to integrate had been, say $\:\sin^3x\cos^4x\,\mathrm dx$, we would have set $$u=\cos x,\quad\mathrm d u=-\sin x\,\mathrm dx.$$

Answer 3

$$\begin{align}\int\sin^4(x)\cos^3(x)\,dx&= - \int-\frac15\cos^2(x)\bigl(-5\cos(x)\sin^4(x)\bigr)\,dx\\ &=- \int-\frac15\cos^2(x)\bigl(\sin^5(x)\bigr)'\,dx. \end{align} $$ Now perform an integration by parts and use the same technique for the new integral.