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

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

3 Answers
Apr 18, 2018

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

Apr 18, 2018

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

Apr 18, 2018

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!