How to integrate intx/(cos^2 (x^2))dxxcos2(x2)dx ?

intx/(cos^2 (x^2))dxxcos2(x2)dx

3 Answers
Apr 18, 2018

see below

Explanation:

We have, int frac{x}[cos^2(x^2)]dxxcos2(x2)dx
Substituting x^2x2 as yy,we get,
2xdx=dy2xdx=dy
Putting this value in the main integral,we get,
frac{1}2intfrac{dy}[ cosy]12dycosy
Or, frac{1}2int"secy " dy12secy dy
Or,frac{ln|tanx^2 +sec x^2|}2 +Clntanx2+secx22+C

Apr 18, 2018

int \frac{x}{cos^2 ( x^2 )} dx = \frac{tan x^2}{2} + Cxcos2(x2)dx=tanx22+C

Explanation:

int \frac{x}{cos^2 ( x^2 )} dx =xcos2(x2)dx=

= \frac{1}{2} int \frac{1}{cos^2 ( x^2 )} (2 x) dx ==121cos2(x2)(2x)dx=

= [ \frac{1}{2} int \frac{1}{cos^2 ( t )} dt ]_{t = x^2} ==[121cos2(t)dt]t=x2=

= [ \frac{1}{2} tan t + C ]_{t = x^2} ==[12tant+C]t=x2=

= \frac{tan x^2}{2} + C=tanx22+C

Apr 18, 2018

The integral is equal to 1/2tan(x^2) + C12tan(x2)+C

Explanation:

Let u = x^2u=x2. Then du = 2x dxdu=2xdx and dx = (du)/(2x)dx=du2x

I = int 1/(2cos^2u) duI=12cos2udu

I = 1/2int sec^2u duI=12sec2udu

This is a known integral

I = 1/2tanu + CI=12tanu+C

Reverse the substitution

I = 1/2tan(x^2) +CI=12tan(x2)+C

Hopefully this helps!