[Math] How to evaluate the trigonometric integral $\int \frac{1}{\cos x+\tan x }dx$

Question

$$\int \dfrac{1}{\cos x+\tan x }dx$$

This can be converted to

$$\int \dfrac{\cos x}{\sin x+\cos^2x}dx$$

But from here I get stuck. Using t substitution will get you into a mess. Are there any tricks which can split the fraction into simpler forms?

Answer 1

HINT:

$$I=\displaystyle\int \dfrac{\cos x}{\sin x+\cos^2x}dx$$

$$=\displaystyle\int \dfrac{\cos x}{\sin x+1-\sin^2x}dx$$

Put $\sin x=u, \cos xdx=du $

$$\implies I=\int \frac{du}{1+u-u^2}=\int \frac{4du}{5-(2u-1)^2}$$

Put $2u-1=v$ and use Partial Fraction Decomposition or $$\int\frac{dx}{a^2-x^2}=\frac1{2a}\ln\left|\frac{a+x}{a-x}\right|+C $$