Question

integrate sinx/sin 3x w.r.t x.

Answer 1

$\int \frac{\sin x}{\sin 3 x}d x=\frac{1}{2 \sqrt{3}} \ln \left(\left|\frac{\sqrt{3}+\tan x}{\sqrt{3}-\tan x}\right|\right)+c$

Given:

$\int \frac{\sin x}{\sin 3 x} d x$

Solution:

$=\int \frac{\sin x}{\sin 3 x} d x$

We know that, $\sin 3 x$

This can be written as

$\sin 3 x=\sin (x+2 x)$

$\sin 3 x=\sin x \cdot \cos 2 x+\cos x+\sin 2 x$

$\rightarrow { since }(\sin(a+b)=\sin a\cdot \cos b+\cos a\cdot \sin b)$

$\sin 3x=\sin x\left( 1-2(\sin x)^{2}+2\sin x(\cos x)^{ 2 } \right.$

$\sin3 x=\sinx-2(\sin x)^{3}+2 \sin x\left(1-(\sin x)^{2}\right)$      

$\sin 3 x=3 \sin x-4 \sin ^{3} x$

Substitute the value in the denominator,

$=\int \frac{\sin x}{3 \sin x-4 \sin ^{3} x} d x$

$=\int\frac { dx }{ 3-4\sin^{ 2 } x }$

$=\int { \frac {dx}{ 3-4\frac { \left( 1-\cos{ 2x}\right)}{2}}}$

$=\int \frac{d x}{3-2(1-\cos 2 x)}$

$=\quad \int { \frac { dx }{ 3-2+2\cos { 2x }}}$

$=\int \frac{d x}{1+2 \cos 2 x}$

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

$Since, \ t=tan x$

$d t=\sec ^{2} x d x$

$=\int { \frac { dt }{ 1+{ t }^{ 2 }+2-2{ t }^{ 2 }}}$

$=\int \frac{d t}{3-t^{2}}$

$\int \frac{\sin x}{\sin 3 x}d x =\frac{1}{2 \sqrt{3}} \ln \left(\left|\frac{\sqrt{3}+\tan x}{\sqrt{3}-\tan x}\right|\right)+c$

Answer 2

Answer:

$\int\frac{\sin x}{\sin3x}\,dx=\frac{\tanh^{-1}\left(\frac{\tan x}{\sqrt3}\right)}{\sqrt3}+C$

Explanation:

As, ${\sin3x}={\sin x(3\cos^2x-\sin^2x)}$

$\frac{\sin x}{\sin3x}=\frac{\sin x}{\sin x(3\cos^2x-\sin^2x)}$

As, ${cos^2x-\sin^2x}={\cos2x}$

$\frac{\sin x}{\sin3x}=\frac{1}{3\cos^2x-\sin^2x}=\frac{1}{2\cos^2x+\cos2x}$

As, ${2cos^2x-1}={\cos2x}$

So, ${2cos^2x}={\cos2x+1}$

$\frac{\sin x}{\sin3x}=\frac{1}{2\cos2x+1}$

As,$cos2x=\frac{1-\tan^2x}{sec^2x}$

So,

$\frac{\sin x}{\sin3x}=\frac{\sec^2x}{3-\tan^2x}$

Let, $u=\tan x,du=\sec^2x\,dx\Rightarrow\int\frac{1}{3-u^2}\,du$

$u=\sqrt3\tanh w,\,du=\sqrt3\text{sech}^2w\,dw,w=\tanh^{-1}\left(\frac{\tan x}{\sqrt3}\right)$

$\int\frac{1}{3-u^2}\,du\Leftrightarrow\frac{1}{\sqrt3}\int\,dw= \frac{w}{\sqrt3}+C$

Thus,

$\int\frac{\sin x}{\sin3x}\,dx=\frac{\tanh^{-1}\left(\frac{\tan x}{\sqrt3}\right)}{\sqrt3}+C$