How do I integrate the $\sin^3\cos^5x$?

Question

How do I find the integral of $$\int\sin^3x\cos^5xdx$$ is it can be solved by reduction formula?

Please explain step by step.

Answer 1

Hint: Are you familiar with $\sin^2x=1-\cos^2x$?

Answer 2

$$\int\sin^3x\cos^5x\mathrm dx$$

Using trig identity, $$\sin^2x=1-\cos^2x$$

We can rewrite $\sin^3$ as

$$=\sin x\cdot\sin^2x$$

$$=\sin x(1-\cos^2x)$$

Thus the integral becomes,

$$\displaystyle\int\sin^3x\cos^5x\mathrm dx=\int\sin x(1-\cos^2x)\cos^5x\mathrm dx$$

Distributing just the cosine, this becomes

$$=\displaystyle\int(\cos^5x-\cos^7x)\sin x\mathrm dx$$

Now use the substitution:

$$u=\cos x \implies \mathrm du=−\sin x\mathrm dx$$

Integrating, this becomes

$$=-\bigg(\dfrac{u^6}{6}-\dfrac{u^8}{8}\bigg)+C$$

Reordering and back-substituting with $u=\cos x$

$$=\dfrac{\cos^8x}{8}-\dfrac{\cos^6x}{6}+C$$

Answer 3

Hint Rewrite the expression as $$(\cos^2x-1)\cos^5x (-\sin x)$$

and recall that $$\frac{d}{dx}f(\cos x)=f'(\cos x)(-\sin x)$$