Question

`int[ tan (log x) + sec^2 (log x)] dx= ` ______

Options

• x sec (log x) + c

• tan (log x) + c

• sec (log x) + c

• x tan (log x) + c

Answer 1

`int[ tan (log x) + sec^2 (log x)] dx= underline (x tan (log x) + c)`

Explanation:

Let `I= int[ tan (log x) + sec^2 (log x)] dx`

Put log x = t

⇒ x = e^t

⇒ dx = e^tdt

`therefore I = int e^t(tan t + sec^2t)dt`

`= e^t tan t + c`   `... [because int e^x [f(x) + f^'(x)dx = e^xf(x)+c]]`

= x tan (log x) + c