Antiderivative of $(\sin^2 x − \sin^3x)^2\cos x$.

Question

Can anyone show me how to antidifferentiate $(\sin^2 x − \sin^3x)^2\cos x$ step by step? I'm a bit confused on what to do.

Answer 1

Since you can observe that the integrand has a function and its derivative, try a $u$-substitution:

$$u = \sin x \quad \mathrm du = \cos x \,\mathrm dx$$

And once you substitute this in, you get:

$$\require{cancel} \int \left(\sin^2 x - \sin^3x\right)^2\cos x \,\mathrm dx = \int \left(u^2-u^3\right)^2\cancel{\cos x} {\mathrm du \over \cancel{\cos x}}$$

You can then expand the polynomial and integrate easily.

Answer 2

$$u = \sin x$$ $$\mathrm du = \cos x$$

$$(\sin^2x-\sin^3x)^2\cos x \,\mathrm dx = (u^2 - u^3)^2\mathrm du $$

$$\int (u^4 - 2u^5 + u^6)\,\mathrm du = \frac15 u^5 - \frac13u^6+\frac17u^7 = \frac15 \sin^5x - \frac13\sin^6x+\frac17\sin^7x + C$$