# Why isn't $\int\sin^2(⁡x)\,dx$ equal to $\frac{\sin^3(x)}3\frac{-1}{\cos(x)}+c$?

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? $$\int\sin⁡^2(x)\,dx =\frac{\sin^3(x)}3\frac{-1}{\cos(x)}+c$$

• I see no step. All I see is a problem and an answer with nothing in between. Cake with no filling is rather boring. – Simply Beautiful Art Dec 7 '16 at 0:38
• Your other method only works for polynomials. Make sure you understand how u-substitution works, and WHY it works for (2x + 1)² – Kaynex Dec 7 '16 at 1:13

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.

Let $u=sin(x)$. This implies that $du = cos(x)dx$ Note that $x = arcsin(u)$, so $cos(x) = \sqrt{1-u^2}$. This makes the integral:

$$\int{u^2 \sqrt{1-u^2}du}$$ which isn't a particularly nice integral either. The reason it works for the example you gave is that you performed a subtitution of $u=2x+1$, so that $du = 2dx$.

In the first case, the subtitution demanded that the substituted quantity be changing nonlinearly with respect to $x$, which didn't result in any sort of desired simplification. In the latter case, the substituted variable changes at a rate of $2$. A factor of $2$ in an integral can be pulled out, as it is a linear operator.

The derivative of the function on the right in your suggested alternative isn't the integrand - because to compute that derivative you need to use the quotient rule and calculate the derivative of $\cos(x)$.

The "basic steps ... for other functions" works in your example because the function inside the parentheses has a constant derivative.