Integrating $\int{3\sin^2x\cos x \;dx}$

Question

I am struggling with the following integral:

$$\int{3\sin^2x\cos x \;dx}$$

I have so far tried to solve this using every tool at my disposal, I have set $t = \sin x$ but I get an even harder quantity to integrate:

$$\int{\frac{t^2-t^4}{\sqrt{1-t^2}} dt}$$

so I have dismissed this way of proceeding. Then I tried to rewrite $\sin^2x = 1-\cos^2x $ but as a result I get stuck at:

$$\int{3dx}-3\int \cos^3xdx$$

which I don't know how to cope with because integrating by parts I end up with an endless chain of integrals. Hope you could shed some light on the way to properly solve this, really feel a bit lost.

Update: sorry I have made a mistake in copying the exercise over the site.

Answer 1

$$\sin^2(x) \cos^2(x) = \dfrac{\sin^2(2x)}4 = \dfrac{1-\cos(4x)}8$$

EDIT

(Question was changed) Setting $\sin(x) = t$. We get that $\cos(x) dx = dt$. Hence, we have $$I = \int 3 \sin^2(x) \cos(x) dx = \int 3 t^2 dt$$

Answer 2

Hint:: $\displaystyle \sin^2x = \frac{1 - \cos 2x}{2}$ and $\displaystyle \cos^2x = \frac{1 + \cos 2x}{2}$

Edit:: Let $\sin x = t $ then $\cos x dx = dt$, then you have $\int 3t^2 dt $

Answer 3

If you let $t=\sin x$, you get $dt = \cos x dx$, so $$\int{3\sin^2x\cos x \;dx}=\int 3t^2dt=t^3+C=\sin^3x+C$$

Answer 4

In the integrand, if all arguments of trig functions are "the x", transform all of them into sin's and cos's. Every m and n below matches any integer.

• cos^2m(x) sin²ⁿ(x) where m and n are nonnegative: discussed below.
• cos²ⁿ⁺¹(x) f(cos²x, sin x): set y = sin x.
• sin²ⁿ⁺¹(x) f(sin²x, cos x): set y = cos x.

If the integrand matches none of these, transform it into tan's and sec's.

• sec²ⁿ(x) f(tan x, sec² x): set y = tan x.

If none of the above work, apply the sneaky Weierstrass substitution. $$ y = \tan \frac x 2 = \frac {\sin x} {\cos x + 1}. $$

Integrate cos^2m(x) sin²ⁿ(x)

Find the less of m and n. Assume n < m, extract sin(2x)²ⁿ with $$ \cos x \sin x = \frac {\sin 2x} 2.$$

There will be either only cos's or only sin's left. Convert them into double-angles with \begin{align} \cos^2 x &= \frac {1 + \cos 2x} 2 \\ \sin^2 x &= \frac {1 - \cos 2x} 2. \end{align}