# How do you find the integral of int sin x * tan x dx?

We need to find the integral of sin x tan x

### Solution

We know that tan x can be expressed as

$\tan x = \frac{\sin x}{\cos x}$ $\int \sin x\tan x dx = \int \sin x \frac{\sin x}{\cos x} dx$ $\int \sin x\tan x dx = \int\sin ^{2}x \sec x$ { 1/cos x = sec x}

$\int \sin x\tan x dx = \int\sec x (1-cos^{2}x)$ $\int \sin x\tan x dx=\int (\sec x -\cos x)dx$ $\int \sin x\tan x dx=\int \sec x -\int \cos x dx$

We know that

$\int \sec x .dx=\ln (\tan x + \sec x)$ $\int\cos x dx= \sin x + C$

Substituting in the above equation we get

$\int \sin x\tan x dx=\ln (\tan x +\sec x)- \sin x + C$

$\int \sin x\tan x dx=\ln (\tan x +\sec x)- \sin x + C$