 Sangeetha Pulapaka
2

\int_{ }^{ }\frac{\sin x\ }{\sin\ x-\cos\ x} dx

Multipying with 2 and dividing with 2 we get,

\frac{1}{2} \int_{ }^{ }\frac{2\sin x}{\sin x\ -\ \cos\ x}

Writing 2 sinx as sinx+sinx +cosx -cosx we get

\frac{1}{2} \int_{ }^{ }\frac{\sin x+\cos x\ +\sin x\ -\cos x}{\sin\ x\ -\ \cos x}

Seperating the first two terms and the last two terms we get

\frac{1}{2}[ \int_{ }^{ }\frac{\sin x+\cos x}{\sin x\ -\ \cos x} dx + \int_{ }^{ }\frac{\sin x\ -\cos x}{\sin\ x\ -\ \cos\ x} dx]

\frac{1}{2} [ \int_{ }^{ }\frac{\sin x+\cos\ x}{\sin\ x\ -\ \cos\ x} dx  + \int_{ }^{ } 1 dx]

Assume sinx - cox = t

Differentiating it we get (cos x +sin x ) dx =  dt (since the differentiation of sinx is                  cosx and the differentiation of cos x = -sin x)

So substituting (sinx +cos x) dx = dt  and sin x - cos x = t

we get \frac{1}{2} \int_{ }^{ }\frac{dt}{t} + \int_{ }^{ }dx

= \frac{1}{2} ln t +\frac{1}{2} x + c

Since t = sinx - cos x

we can write this as \frac{1}{2} ln | sinx - cos x| + c , where c is the constant of integration. Vivekanand Vellanki
2

Integrals with trigonometric entities becomes easy if you can make the denominator 1.

The answer given by Sangeetha did that - in a very non-intuitive way.

Another approach is the following:

\frac{\sin x}{\sin x-\cos x}=\frac{\sin x\left(\sin x+\cos x\right)}{\sin^2x-\cos^2x}=\frac{\left(\sin^2x+\sin x\cos x\right)}{\sin^2x-\cos^2x}

Now, using the following:

\cos2x=\cos^2x-\sin^2x and

\sin2x=2\sin x\cos x and

\sin^2x=\frac{\left(1-\cos2x\right)}{2}

This gives the following:\frac{\left(\sin^2x+\sin x\cos x\right)}{\sin^2x-\cos^2x}=\frac{\left(\frac{\left(1-\cos2x\right)}{2}+\frac{1}{2}\sin2x\right)}{-\cos2x}=\frac{-\frac{1}{2}\left(1-\cos2x+\sin2x\right)}{\cos2x}

\frac{\left(-\frac{1}{2}\left(1-\cos2x+\sin2x\right)\right)}{\cos2x}=-\frac{1}{2}\left(\sec2x-1+\tan2x\right)

Now, there is no denominator and it should be possible to integrate this expression.