Question

Integrate $\int \frac{\sin \left(\mathrm{x}\right)}{2+\sin \left(2\mathrm{x}\right)}$.

Answer 1

Step-1 Converting in integrable form:

Let, $I=\int \frac{\sin \left(x\right)}{2+\sin \left(2x\right)}\mathrm{d}x$

$\begin{array}{ccc}\Rightarrow & I=\int \frac{\sin \left(x\right)}{1+{\sin }^2\left(x\right)+{\cos }^2\left(x\right)+2\sin \left(x\right)\cos \left(x\right)}\mathrm{d}x& \left[\because {\sin }^2\left(x\right)+{\cos }^2\left(x\right)=1,\sin \left(2x\right)=2\sin \left(x\right)\cos \left(x\right)\right]\\ \Rightarrow & I=\int \frac{\sin \left(x\right)}{1+{\left[\sin \left(x\right)+\cos \left(x\right)\right]}^2}\mathrm{d}x& \left[\because {\left(a+b\right)}^2=a^2+b^2+2\mathrm{ab}\right]\end{array}$

Add $\sin \left(x\right)$ to the numerator and divide by $2$

$\Rightarrow I=\frac{1}{2}\int \frac{\sin \left(x\right)+\sin \left(x\right)}{1+{\left[\sin \left(x\right)+\cos \left(x\right)\right]}^2}$

Add and subtract $\cos \left(x\right)$ from the numerator

$$\begin{gathered} \Rightarrow I=\frac{1}{2}\int \frac{\sin \left(x\right)+\sin \left(x\right)+\cos \left(x\right)-\cos \left(x\right)}{1+{\left[\sin \left(x\right)+\cos \left(x\right)\right]}^2}\mathrm{d}x \\ \Rightarrow I=\frac{1}{2}\int \frac{\sin \left(x\right)-\cos \left(x\right)}{1+{\left[\sin \left(x\right)+\cos \left(x\right)\right]}^2}\mathrm{d}x+\frac{\sin \left(x\right)+\cos \left(x\right)}{1+{\left[\sin \left(x\right)+\cos \left(x\right)\right]}^2}\mathrm{d}x \end{gathered}$$

Let $u=\frac{1}{2}\int \frac{\sin \left(x\right)-\cos \left(x\right)}{1+{\left[\sin \left(x\right)+\cos \left(x\right)\right]}^2}\mathrm{d}x$ and $v=\frac{1}{2}\int \frac{\sin \left(x\right)+\cos \left(x\right)}{1+{\left[\sin \left(x\right)+\cos \left(x\right)\right]}^2}\mathrm{d}x$

$\Rightarrow I=u+v$

Step-2 : Integration by substitution of $u$:

For $u$, let $\sin \left(x\right)+\cos \left(x\right)=t$

Differentiating both sides with respect to $x$,

$\begin{array}{ccc}\Rightarrow & \cos \left(x\right)-\sin \left(x\right)\mathrm{d}x=\mathrm{d}t& \\ \Rightarrow & u=-\frac{1}{2}\int \frac{\mathrm{d}t}{1+t^2}& \\ \Rightarrow & u=-\frac{{\tan }^{-1}\left(t\right)}{2}+C& [\because \int \frac{1}{1+x^2}={\tan }^{-1}\left(x\right)]\\ \Rightarrow & u=-\frac{{\tan }^{-1}\left[\sin \left(x\right)+\cos \left(x\right)\right]}{2}+C& \end{array}$

Step-2 : Integration by substitution of $v$:

We know that $1+{\left[\sin \left(x\right)+\cos \left(x\right)\right]}^2=2+\sin \left(2x\right)$ (we derived this at the start)

So, $v=\int \frac{\sin \left(x\right)+\cos \left(x\right)}{2+\sin \left(2x\right)}\mathrm{d}x$

Add and subtract $1$ from denominator

$\begin{array}{ccc}\Rightarrow & \mathrm{v}=\frac{1}{2}\int \frac{\sin \left(x\right)+\cos \left(x\right)}{2+1+\sin \left(2x\right)-1}\mathrm{d}x& \\ \Rightarrow & \mathrm{v}=\frac{1}{2}\int \frac{\sin \left(x\right)+\cos \left(x\right)}{3-\left[1-\sin \left(2x\right)\right]}\mathrm{d}x& \\ \Rightarrow & \mathrm{v}=\frac{1}{2}\int \frac{\sin \left(x\right)+\cos \left(x\right)}{3-\left[{\sin }^2\left(x\right)+{\cos }^2\left(x\right)-2\sin \left(x\right)\cos \left(x\right)\right]}\mathrm{d}x& \left[\because {\sin }^2\left(x\right)+{\cos }^2\left(x\right)=1,\sin \left(2x\right)=2\sin \left(x\right)\cos \left(x\right)\right]\\ \Rightarrow & \mathrm{v}=\frac{1}{2}\int \frac{\sin \left(\mathrm{x}\right)+\cos \left(\mathrm{x}\right)}{3-{\left[\sin \left(\mathrm{x}\right)-\cos \left(\mathrm{x}\right)\right]}^2}\mathrm{d}x& \left[\because {\left(a-b\right)}^2=a^2+b^2-2\mathrm{ab}\right]\end{array}$

Let $\sin \left(x\right)-\cos \left(x\right)=t$

Differentiate both sides with respect to $x$,

$\Rightarrow \sin \left(x\right)-\cos \left(x\right)\mathrm{d}x=\mathrm{d}t$

$\begin{array}{ccc}\Rightarrow & \mathrm{v}=\frac{1}{2}\int \frac{\mathrm{d}t}{3-t^2}& \\ \Rightarrow & \mathrm{v}=\frac{1}{4\sqrt{3}}\log \left(\frac{\sqrt{3}-t}{\sqrt{3}+t}\right)& \left[\because \int \frac{\mathrm{dt}}{a^2-t^2}=\frac{1}{2a}\log \left(\frac{a-t}{a+t}\right)\right]\\ \Rightarrow & \mathrm{v}=\frac{1}{4\sqrt{3}}\log \left(\frac{\sqrt{3}-\left[\sin \left(\mathrm{x}\right)-\cos \left(\mathrm{x}\right)\right]}{\sqrt{3}+\left[\sin \left(\mathrm{x}\right)-\cos \left(\mathrm{x}\right)\right]}\right)+\mathrm{C}& \end{array}$

But, $I=u+v$

$\Rightarrow I=-\frac{{\tan }^{-1}\left[\sin \left(x\right)+\cos \left(x\right)\right]}{2}+\frac{1}{4\sqrt{3}}\log \left(\frac{\sqrt{3}-\left[\sin \left(x\right)-\cos \left(x\right)\right]}{\sqrt{3}+\left[\sin \left(x\right)-\cos \left(x\right)\right]}\right)+C$

Therefore, $\mathbf{\int }\frac{\mathbf{sin}\left(x\right)}{\mathbf{2}\mathbf{+}\mathbf{sin}\left(2x\right)}\mathbf{=}\mathbf{-}\frac{{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left[\sin \left(x\right)+\cos \left(x\right)\right]}{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{4}\sqrt{\mathbf{3}}}\mathbf{log}\left(\frac{\sqrt{3}-\left[\sin \left(x\right)-\cos \left(x\right)\right]}{\sqrt{3}+\left[\sin \left(x\right)-\cos \left(x\right)\right]}\right)\mathbf{+}\mathbf{C}$.