Question

Solve $ \int \dfrac{{\sin x}}{{\left( {\sin x - \cos x} \right)}}dx $

Answer 1

Hint: Here first we are rationalizing the denominator by multiplying the numerator and denominator with the denominator’s conjugate. Conjugate of the denominator is $ \left( {\sin x + \cos x} \right) $ . And then we get a result in squares of the terms. So use the below mentioned formulas appropriately and find the integration value.
Formulas used:
1. $ {\cos ^2}x - {\sin ^2}x = \cos 2x $
2. $ \sin 2x = 2\sin x\cos x $
3. $ {\sin ^2}x = \dfrac{{1 - \cos 2x}}{2} $
4. $ \int \tan xdx = \log \left| {secx} \right| + c $
5. $ \int secxdx = \log \left| {secx + \tan x} \right| + c $

Complete step-by-step answer:
We are given to solve the integration $ \int \dfrac{{\sin x}}{{\left( {\sin x - \cos x} \right)}}dx $
Multiplying the numerator and denominator of the above trigonometric expression by denominator’s conjugate $ \left( {\sin x + \cos x} \right) $ , we get $ \int \dfrac{{\sin x\left( {\sin x + \cos x} \right)}}{{\left( {\sin x - \cos x} \right)\left( {\sin x + \cos x} \right)}}dx $
Considering $ \sin x $ as ‘a’ and $ \cos x $ as ‘b’, the denominator will be in the form $ \left( {a - b} \right)\left( {a + b} \right) $ which is equal to $ {a^2} - {b^2} $
Therefore, the expression becomes $ \int \dfrac{{\sin x\left( {\sin x + \cos x} \right)}}{{\left( {{{\sin }^2}x - {{\cos }^2}x} \right)}} $
Expanding the numerator, we get
 $ \Rightarrow \int \dfrac{{{{\sin }^2}x + \sin x\cos x}}{{\left( {{{\sin }^2}x - {{\cos }^2}x} \right)}}dx $
The numerator is now a sum of two terms, so the integration can be divided into two smaller integrations
 $ \Rightarrow \int \dfrac{{{{\sin }^2}x}}{{\left( {{{\sin }^2}x - {{\cos }^2}x} \right)}}dx + \int \dfrac{{\sin x\cos x}}{{\left( {{{\sin }^2}x - {{\cos }^2}x} \right)}}dx $
We know that $ {\cos ^2}x - {\sin ^2}x = \cos 2x $ and we have $ {\sin ^2}x - {\cos ^2}x $ in the denominator of the above integrations. So the value of $ {\sin ^2}x - {\cos ^2}x $ is $ - \cos 2x $ . Therefore the integrations become
 $ \Rightarrow \int \dfrac{{{{\sin }^2}x}}{{ - \cos 2x}}dx + \int \dfrac{{\sin x\cos x}}{{ - \cos 2x}}dx $
And $ {\sin ^2}x = \dfrac{{1 - \cos 2x}}{2} $
 $ \Rightarrow - \int \dfrac{{\left( {\dfrac{{1 - \cos 2x}}{2}} \right)}}{{\cos 2x}}dx - \int \dfrac{{\sin x\cos x}}{{\cos 2x}}dx = - \int \dfrac{{1 - \cos 2x}}{{2\left( {\cos 2x} \right)}}dx - \int \dfrac{{\sin x\cos x}}{{\cos 2x}}dx $
The value of $ \sin 2x = 2\sin x\cos x $ , this gives the value of $ \sin x\cos x = \dfrac{{\sin 2x}}{2} $
 $ \Rightarrow - \int \dfrac{{1 - \cos 2x}}{{2\left( {\cos 2x} \right)}}dx - \int \dfrac{{\left( {\dfrac{{\sin 2x}}{2}} \right)}}{{\cos 2x}}dx = - \int \dfrac{{1 - \cos 2x}}{{2\left( {\cos 2x} \right)}}dx - \int \dfrac{{\sin 2x}}{{2\cos 2x}}dx $
Taking $ \dfrac{1}{2} $ common out, we get
 $ \Rightarrow \dfrac{1}{2}\left[ { - \int \dfrac{{1 - \cos 2x}}{{\cos 2x}}dx - \int \dfrac{{\sin 2x}}{{\cos 2x}}dx} \right] $
Expanding the first integration into two, we get
 $ \Rightarrow \dfrac{1}{2}\left[ { - \int \dfrac{1}{{\cos 2x}}dx + \int \dfrac{{\cos 2x}}{{\cos 2x}}dx - \int \dfrac{{\sin 2x}}{{\cos 2x}}dx} \right] $
The inverse of cosine is secant and the ratio of sine and cosine is tangent, then we get
 $ \Rightarrow \dfrac{1}{2}\left[ { - \int sec2xdx + \int 1dx - \int \tan 2xdx} \right] $
We know that $ \int secxdx = \log \left| {secx + \tan x} \right| + c $ and $ \int \tan xdx = \log \left| {secx} \right| + c $ , but here we have 2x in the place of x.
Therefore, we get
 $ \Rightarrow \dfrac{1}{2}\left[ { - \left( {\dfrac{{\log \left| {sec2x + \tan 2x} \right|}}{2}} \right) + x - \dfrac{{\log \left| {sec2x} \right|}}{2} + c} \right] $
Sending $ \dfrac{1}{2} $ inside, we get
 $ \Rightarrow - \left( {\dfrac{{\log \left| {sec2x + \tan 2x} \right|}}{4}} \right) + \dfrac{x}{2} - \dfrac{{\log \left| {sec2x} \right|}}{4} + c $
Therefore, the value of $ \int \dfrac{{\sin x}}{{\left( {\sin x - \cos x} \right)}}dx $ is $ - \left( {\dfrac{{\log \left| {sec2x + \tan 2x} \right|}}{4}} \right) + \dfrac{x}{2} - \dfrac{{\log \left| {sec2x} \right|}}{4} + c $
So, the correct answer is “$ - \left( {\dfrac{{\log \left| {sec2x + \tan 2x} \right|}}{4}} \right) + \dfrac{x}{2} - \dfrac{{\log \left| {sec2x} \right|}}{4} + c $ ”.

Note: ‘c’ is known as the integration constant and the sum of constants is always a constant. So we can write only one ‘c’ for the whole integration. And lastly we have divided the integration results by 2 because, when any trigonometric function has an angle like ‘nx’, the result must be divided by n. Here the angle is 2x, so we have divided by 2. Conjugate of an expression will have a different sign from the original.