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

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

Answer

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

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