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.


4 Answers 4


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


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

  • $\begingroup$ Thank you for your quick answer, my question will probably sound stupid, but I'd like to ask why is that $dt = \cos(x)dx$ and don't we set $\cos(x) = \sin^{-1} t$ and then calculate the $dt$ value by derivating $\sin^{-1}$? $\endgroup$
    – Mary
    Apr 17, 2013 at 17:51
  • $\begingroup$ @Mary I assume you mean $x = \sin^{-1}(t)$ and then working out $dx$ from that. That is an inefficient way to go about. If you have $t = \sin(x)$, then $\dfrac{dt}{dx} = \cos(x) \implies dt = \cos(x) dx$ $\endgroup$
    – user17762
    Apr 17, 2013 at 18:04

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 $


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


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.

  • cos2m(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 cos2m(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}


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