# Evaluate the following integrals:To solve this type of solution, we are going to substitute the value of $sin x$ and $cos x$ in terms of $an (x / 2)$

Question:

Evaluate $\int \frac{1}{\sin x+\sin 2 x} d x$

Solution:

To solve this type of solution, we are going to substitute the value of $\sin x$ and $\cos x$ in terms of $\tan (x / 2)$

$\sin x=\frac{2\left[\tan \frac{x}{2}\right]}{1+\tan ^{2} \frac{x}{2}}$

$\cos x=\frac{\left(1-\frac{\tan ^{2} x}{2}\right)}{1+\frac{\tan ^{2} x}{2}}$

$I=\int \frac{1}{\frac{2 \tan x / 2}{1+\tan ^{2} \frac{x}{2}}\left(1+2 \cdot \frac{\left.1-\tan ^{2} \frac{x}{2}\right)}{1+\tan \frac{2}{2}}\right)} d x$

$I=\int \frac{\sec ^{2} \frac{x}{2}}{2 \tan \frac{x}{2}\left(3-\tan ^{2} \frac{x}{2}\right)} d x$

In this type of equations we apply substitution method so that equation may be solve in simple way

Let $\tan \left(\frac{x}{2}\right)=t$

$\frac{1}{2} \cdot \sec ^{2} \frac{x}{2} d x=d t$

Put these terms in above equation, we get $I=\int \frac{d t}{t\left(3-t^{2}\right)}$

$I=\int \frac{t^{-3} d t}{\left(3 t^{-2}-1\right)}$

Let us now again apply the substitution method in above equation

Let $\mathrm{t}^{-2}=\mathrm{k}$

$-2 \cdot t^{-3} d t=d k$

Substitute these terms in above equation gives-

$I=-\frac{1}{6} \int \frac{d k}{k}$

$I=\frac{1}{6 k^{2}}$

$=\frac{1}{6} \cdot\left(\frac{3-t^{2}}{t^{2}}\right)^{2}$

$=\frac{1}{6} \cdot\left(\frac{3}{t^{2}}-1\right)^{2}$

Now put the value of $t, t=\tan (x / 2)$ in above equation gives us the finally solution

$I=\frac{1}{6} \cdot\left(\frac{3}{\tan ^{2} \frac{x}{2}}-1\right)^{2}$