Question

How do you integrate $\int \tan xdx$.

Answer 1

Step 1: Assume a variable

The given integration is,

$\int \tan xdx=\int \frac{\sin x}{\cos x}dx$

Let $u=\cos x$.

$\therefore du=-\sin xdx$

Step 2: Evaluate the integration

Substituting this in the above equation we get,

$I=-\int \frac{du}{u}$

$=-\ln u+c$ $[\because \int \frac{1}{x}dx=ln\left|x\right|,\text{c be the integration constant}]$

$=-\ln \left(\left|\cos x\right|\right)+c$

$=\ln \left(\left|\frac{1}{\cos x}\right|\right)+c$

$=\ln \left(\left|secx\right|\right)+c$

Hence, the required answer is $\mathit{l}\mathit{n}\left(\left|secx\right|\right)\mathbf{+}\mathit{c}$.