Find integral of the following?

Question

#intcos^2x dx#

1346 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-21T17:21:24

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

#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)]#

Answer 2 · 2018-04-21T17:24:12

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

Here,

#I=intcos^2xdx#

#=int(1+cos2x)/2#

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

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