Question

Evaluate: $\int \frac{1}{(sinx+tanx)}dx$

Answer 1

$LetI=\int \frac{1}{(\sin x+\tan x)}dxI=\int \frac{1}{(\sin x+\frac{\sin x}{\cos x})}dxI=\int \frac{1}{\left(\frac{\sin x\cos x+\sin x}{\cos x}\right)}dx=\int \frac{\cos x}{\sin x\cos x+\sin x}dx=\int \frac{\cos x}{\sin x(\cos x+1)}dx=\int \frac{\sin x\cos x}{{\sin }^{2}x(\cos x+1)}dx=\int \frac{\sin x\cos x}{(1-{\cos }^{2}x)(\cos x+1)}dx=\int \frac{\sin x\cos x}{(1-\cos x){(\cos x+1)}^2}dx$

Let $u=\cos x⟹\frac{du}{dx}=-\sin x$

Substituting into our integral, we get:

$I=\int \frac{(-1)\left(u\right)}{(1-u){(u+1)}^2}du=\int \frac{u}{(u-1){(u+1)}^2}du$

which we can decompose into partial fraction as:

$\frac{u}{(u-1){(u+1)}^2}=\frac{A}{u-1}+\frac{B}{u+1}+\frac{C}{{(u+1)}^2}=\frac{A{(u+1)}^2+B(u-1)(u+1)+C(u-1)}{(u-1){(u+1)}^2}$

leading to,

$u=A{(u+1)}^2+B(u-1)(u+1)+C(u-1)$

Put $u=1⟹1=4A⟹A=\frac{1}{4}$

Put $u=-1⟹-1=-2C⟹C=\frac{1}{2}$

Comparing coefficients,

$coeff\left(u^2\right)⟹0=A+B⟹B=-\frac{1}{4}$

Hence, the partial fraction decomposition gives us:

$I=\int \frac{\frac{1}{4}}{u-1}-\frac{\frac{1}{4}}{u+1}+\frac{\frac{1}{2}}{{(u+1)}^2}du=\frac{1}{4}\int \frac{1}{u-1}du-\frac{1}{4}\int \frac{1}{u+1}du+\frac{1}{2}\int \frac{1}{{(u+1)}^2}du=\frac{1}{4}\ln |u-1|-\frac{1}{4}\ln |u+1|+\frac{1}{2}\frac{-1}{u+1}+C=\frac{1}{4}\ln \left|\frac{u-1}{u+1}\right|-\frac{1}{2}\frac{1}{u+1}+C$

Restoring the substitution, we get:

$I=\frac{1}{4}\ln \left|\frac{\cos x-1}{\cos x+1}\right|-\frac{1}{2}\frac{1}{\cos x+1}+C$