Find integral of the following?

#intcos^2x dx#

2 Answers
Apr 21, 2018

the answer #1/2[x+1/2sin(2x)]+c#

Explanation:

#intcos^2(x)*dx#

#1/2int(1+cos(2x))*dx#

#1/2[x+1/2sin(2x)]+c#

Note That:

#cos^2(x)=1/2[1+cos(2x)]#

#sin^2(x)=1/2[1-cos(2x)]#

Apr 21, 2018

#I=1/4(2x+sin2x)+c#

Explanation:

Here,

#I=intcos^2xdx#

#=int(1+cos2x)/2#

#=1/2[x+sin(2x)/2]+c#

#=1/4(2x+sin2x)+c#