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!