Question

How to integrate `intx/(cos^2 (x^2))dx` ?

`intx/(cos^2 (x^2))dx`

Answer 1

see below

Explanation:

We have, `int frac{x}[cos^2(x^2)]dx`
Substituting `x^2` as `y`,we get,
`2xdx=dy`
Putting this value in the main integral,we get,
`frac{1}2intfrac{dy}[ cosy]`
Or, `frac{1}2int"secy " dy`
Or,`frac{ln|tanx^2 +sec x^2|}2 +C`

Answer 2

`int \frac{x}{cos^2 ( x^2 )} dx = \frac{tan x^2}{2} + C`

Explanation:

`int \frac{x}{cos^2 ( x^2 )} dx =`

`= \frac{1}{2} int \frac{1}{cos^2 ( x^2 )} (2 x) dx =`

`= [ \frac{1}{2} int \frac{1}{cos^2 ( t )} dt ]_{t = x^2} =`

`= [ \frac{1}{2} tan t + C ]_{t = x^2} =`

`= \frac{tan x^2}{2} + C`

Answer 3

The integral is equal to `1/2tan(x^2) + C`

Explanation:

Let `u = x^2`. Then `du = 2x dx` and `dx = (du)/(2x)`

`I = int 1/(2cos^2u) du`

`I = 1/2int sec^2u du`

This is a known integral

`I = 1/2tanu + C`

Reverse the substitution

`I = 1/2tan(x^2) +C`

Hopefully this helps!