Using the trigonometry "half-angle" identity $\sin^2(x)=\frac12(1-\cos(2x))$:
$$\int\sin^2(x)\,dx
=\int\left(\frac12-\frac12\cos(2x)\right)\,dx
=\frac12x-\frac14\sin(2x)+c$$
I know below method is wrong. But following the basic steps of integral that can be used for other functions such as $(2x+1)^2$, why doesn't this step work?
$$\int\sin^2(x)\,dx
=\frac{\sin^3(x)}3\frac{-1}{\cos(x)}+c$$
Best Answer
Simply put: it doesn't work because it doesn't work. If you try to differentiate that expression, you will not get $\sin^2(x)$. One can use a sort of "reverse chain rule" if you will, on functions like $(2x+1)^2$ to get an antiderivative as $\frac16(2x+1)^3$, because the derivative of the interior function is constant. When the interior function has a derivative which depends on $x$, this is not so easy (by interior function in this case I mean $2x+1$). If the function is $(2x^2+1)^2$, this becomes more difficult to use the "reverse chain rule" method to find an antiderivative. You cannot simply divide by the interior function's derivative now (in the antiderivative), because it depends on $x$, and if you divide by a function of $x$, that changes the derivative entirely. So in the case of $\sin(x)$, which has a derivative which depends on $x$, we cannot use this "reverse chain rule".
For $f(x)=(2x+1)^2$, we can use the "reverse chain rule" by applying a substitution (whether explicitly or implicitly). By letting $u=2x+1$ we see that $du=2dx$, so: $$\int f(x)\,dx=\int (2x+1)^2\,dx=\int (u)^2\frac12\,du=\frac13u^3+c$$ Now in the case of $g(x)=\sin^2(x)$, we see that the interior function, $\sin(x)$, has a non-constant derivative, $\cos(x)$. So to make a substitution for the interior function, we would get $u=\sin(x)$ so $du=cos(x)dx$, and then: $$\int g(x)\,dx=\int\sin^2(x)\,dx=\int u^2\frac1{\cos(x)}\,du$$ which makes things worse really.