Question

How do you find the integral of ${{\sin }^{2}}(ax)$?

Answer 1

Hint: In this problem we will use the one of the trigonometric formulas that is $\cos 2x=1-2{{\sin }^{2}}x$. From this formula we will write the value of ${{\sin }^{2}}(ax)$. Now we will distribute the denominator to individual terms in the numerator. After that, we will apply the integration to each term individually and use the appropriate integration formulas to obtain the result.

Formulas Used:
1. $\cos 2x=1-2{{\sin }^{2}}x$
2.$\int{\cos axdx=\dfrac{\sin ax}{a}+C}$
3.$\int{\cos xdx=\sin x}+C$
4.$\int{1.dx=x}+C$
5.$\sin 2x=2\sin x\cos x$

Complete Step by Step Procedure:
Given that,${{\sin }^{2}}(ax)$
From the trigonometric formula $\cos 2x=1-2{{\sin }^{2}}x$, the value of ${{\sin }^{2}}x$ is $\dfrac{1-\cos x}{2}$. From this we are writing ${{\sin }^{2}}ax$ as
$\Rightarrow {{\sin }^{2}}(ax)=\dfrac{1-\cos (2ax)}{2}$
Applying integration on both sides of the above equation, then we will get
$\int{{{\sin }^{2}}(ax)dx=\int{\dfrac{1-\cos (2ax)}{2}}}dx$
Distributing the denominator over the numerator in RHS, then we will get
$\int{{{\sin }^{2}}(ax)dx=\int{\left( \dfrac{1}{2}-\dfrac{\cos 2ax}{2} \right)dx}}$
Applying the integration to each term individually, then we will get
$\begin{align}
  & \Rightarrow \int{{{\sin }^{2}}axdx}=\int{\dfrac{1}{2}dx-\int{\dfrac{\cos 2ax}{2}dx}} \\
 & \Rightarrow \int{{{\sin }^{2}}axdx}=\dfrac{1}{2}\int{1.dx}-\dfrac{1}{2}\int{\cos 2axdx} \\
\end{align}$
Now we using the formula $\int{1.dx=x}+C$ in the above equation, then we will get
$\dfrac{1}{2}\int{1.dx}=\dfrac{x}{2}+C$
Substituting this value in the integration value, then we will get
$\Rightarrow \int{{{\sin }^{2}}axdx}=\dfrac{x}{2}-\dfrac{1}{2}\int{\cos 2axdx}+C$
Using the formula $\int{\cos axdx=\dfrac{\sin ax}{a}+C}$ in the above equation, then we will get
$\Rightarrow \int{{{\sin }^{2}}ax}dx=\dfrac{x}{2}-\dfrac{1}{2}\left( \dfrac{\sin 2ax}{2a} \right)+C$
Simplifying the above equation, then we will get
$\Rightarrow \int{{{\sin }^{2}}ax}dx=\dfrac{x}{2}-\dfrac{1}{4a}\sin 2ax+C$
We know that $\sin 2x=\sin x\cos x$. Substituting this value in the above equation, then we will get
$\Rightarrow \int{{{\sin }^{2}}ax}dx=\dfrac{x}{2}-\dfrac{1}{4a}\left[ 2\sin ax\cos ax \right]+C$
Now we will simplify the above equation then
$\Rightarrow \int{{{\sin }^{2}}ax}dx=\dfrac{x}{2}-\dfrac{\sin ax\cos ax}{2a}+C$
Taking the LCM for the first two terms in the RHS, then we will get

$\Rightarrow \int{{{\sin }^{2}}ax}dx=\dfrac{ax-\sin ax\cos x}{2a}+C$

Note:
For this problem we have the result after integration as $\int{{{\sin }^{2}}ax}dx=\dfrac{x}{2}-\dfrac{1}{4a}\sin 2ax+C$ . Students may stop their process after getting this result. But the problem is not finished. We have to use the formula $\sin 2x=\sin x\cos x$ and we need to simplify further. So, don’t stop the solution after integration, try to simplify the obtained result by using the well-known formulas.