How to evaluate the following integral ? What substitution will be helpful?

Question

$$\int\frac{cotx }{(1-sinx)(secx+1)}dx $$ we can write this as $$\int \frac {cosecx.cotx}{(cosecx-1)(secx+1)}dx $$ Now $cosecx=t$ gives $cosecx.cotx=-dt $ which appears in the numerator, what to do about $secx+1$ ?

Answer 1

$\displaystyle\int\frac{cotx }{(1-sinx)(secx+1)}dx =\int\frac{cos^2x }{sinx(1-sinx)(1+cosx)}dx =\int\frac{1+sinx}{sinx(cosx+1)}dx$

$\displaystyle\int\frac{1+sinx}{sinx(cosx+1)}dx=\int\frac{1}{sinx(cosx+1)}dx+\int\frac{1}{(cosx+1)}dx$

$\displaystyle\int\frac{1}{4sin(x/2)cos^3(x/2)}dx+\int\frac{sec^2(x/2)}{2}dx$

$\displaystyle cos(x/2)=t \Rightarrow dx=\frac{2dt}{-sin(x/2)}$

$\displaystyle -\int\frac{1}{2(1-t^2)\cdot t^3}dt+\int\frac{sec^2(x/2)}{2}dx$

Now usig partial fractions we get ,

$\displaystyle \int\frac{1}{2(1-t^2)\cdot t^3}dt= \frac{-1}{4 t^2} + \frac{ln[t]}{2} - \frac{ln[1 - t^2]}{4}$

$\displaystyle -\int\frac{1}{2(1-t^2)\cdot t^3}dt+\int\frac{sec^2(x/2)}{2}dx=$$\displaystyle \frac{1}{4 cos^2(x/2)} - \frac{ln[cos(x/2)]}{2}+ \frac{ln[sin^2(x/2)]}{4}+tan^2(x/2)+C$

Answer 2

$$\frac{\cot x}{(1-\sin x)(1+\sec x)}=\frac{1+\sin x}{\sin x(1+\cos x)}=\frac{1}{\sin x(1+\cos x)}+\frac{1}{1+\cos x}$$

Enforcing the Wiereatrass Substitution $\tan(x/2)=u$ and $dx=\frac{2}{u^2+1}\,du$ reveals

$$\begin{align} \int \frac{\cot x}{(1-\sin x)(1+\sec x)}\,dx&=\int \left(\frac{(u^2+1)^2}{4u}+\frac{u^2+1}{2}\right)\,\frac{2}{u^2+1}du\\\\ &=\int \left(\frac12 u +\frac12 u^{-1}+1\right)\,du\\\\ &=\frac14 \tan^2\left(\frac{x}{2}\right)+\frac12 \log\left(\tan \left(\frac{x}{2}\right)\right)+\tan\left(\frac{x}{2}\right)+C \end{align}$$

Answer 3

The function can be simplified to $$ \frac{1}{1+\cos x}+\frac{\csc x}{1+\cos x}. $$ Since $$ \int\frac{1}{1+\cos x}\,dx=\frac{\sin x}{1+\cos x}+C, $$ we only need to take care of the second term. Write it as $$ \frac{\sin x}{(1-\cos^2x)(1+\cos x)} $$ and let $u=\cos x$ which will give you $$ -\int\frac{1}{(1-u^2)(1+u)}\,du $$ which I'm sure you can handle with partial fraction decomposition. You can compare the final result (of your whole thing) with what I get: $$ \frac{\sin x}{1+\cos x}+\frac{1}{2(1+\cos x)}+\frac{1}{4}\log\frac{1-\cos x}{1+\cos x}+C. $$