Integration of $\sin^2x$ without using double angle identity of $\cos 2x$ I want to integrate $\sin^2x$ without using the double angle identity of $\cos2x$.
Here's what I tried:
$$
\int \sin^2x dx
= \int \sin x \tan x \cos x dx
$$
Let $u = \sin x$ => $du = \cos x dx$
And if $u = \sin x$, $\tan x = \frac{u}{\sqrt{1-u^2}}$, therefore
$$
=\int u × \frac{u}{\sqrt{1-u^2}}du = \int \frac{u^2}{\sqrt{1-u^2}}du
$$
Now if $t = \sqrt{1 - u^2}$, $2dt = \frac{du}{\sqrt{1-u^2}}$ and $u^2 = 1 - t^2$
$$
= 2\int (1-t^2)dt = 2t - \frac{2t^3}{3}$$
Substituting back $u$ and $\sin x$
$$
= 2\sqrt{1-u^2} - \frac{2}{3}(\sqrt{1 - u^2})^3
$$
$$
= 2\cos x - \frac{2}{3}\cos^3x
$$
But when you differentiate it you get $-2sin^3x$
All the steps seem right to me, why is the answer wrong or what I did is wrong, and is using the double angle formula the only way to integrate it?
 A: You can find both $$\int \sin^2xdx$$ and $$\int \cos^2 x dx $$ at the same time using the following.
Apply integration by parts on $$\int \sin^2xdx$$ using $u=\sin x $ and $dv=\sin x dx $ to get $$\int \sin^2 xdx= -\sin x \cos x + \int \cos^2 xdx$$
Thus we have a system $$ \int \sin^2xdx-\int \cos^2 xdx= -\sin x \cos x$$
$$ \int \sin^2xdx+\int \cos^2 x dx = \int 1 dx =x$$
Upon adding and subtracting the two above formulas you will get the answer for both$$ \int \sin^2 xdx=(1/2)(x-\sin x \cos x)+C$$ and  $$\int \cos^2x dx=(1/2)(x+\sin x \cos x)+C$$
A: You can avoid using the double angle formula by integrating by parts $$\int\sin^2x\,\mathrm{d}x = -\cos x \sin x + \int \cos^2 x \,\mathrm{d}x = -\cos x \sin x + x - \int \sin^2 x \,\mathrm{d}x. $$
A: You forgot to apply the chain rule:  $$t=\sqrt{1-u^2}\implies dt=-\frac{u}{\sqrt{1-u^2}}\,du.$$
A: If $t=\sqrt{1-u^2}$, then$$\mathrm dt=-\frac u{\sqrt{1-u^2}}\,\mathrm du\quad\text{and}\quad u^2=1-t^2.$$So,\begin{align}\int\frac{u^2}{\sqrt{1-u^2}}\,\mathrm du&=-\int u\frac{-u}{\sqrt{1-u^2}}\,\mathrm du\\&=-\int\sqrt{1-t^2}\,\mathrm dt\\&=-\frac12\left(t\sqrt{1-t^2}+\arcsin (t)\right).\end{align}
A: People have already mentioned that $\frac{dt}{du}$ = -$\frac{u}{t}$ but if you follow that line your integral then becomes:
$\int \frac{1-t^2}{t^2} \sqrt{1-t^2} dt$
Which may not be any prettier to solve. As an alternative I suggest using Euler's identity to solve this.
$sin(x) = \frac{e^{ix}-e^{-ix}}{2i}$
It'll be much quicker to get to the answer with much more trivial integrals to solve.
