Question #642d1

2 Answers
Jan 17, 2018

#int cos^2xsin^2x dx = (4x-sin4x)/32 +C#

Explanation:

Using the trigonometric identity:

#sin2x = 2sinxcosx#

we have that:

#cos^2xsin^2x = (sin^2 2x)/4#

so:

#int cos^2xsin^2x dx = 1/4 int sin^2 (2x)dx #

Use now:

#sin^2 2x = (1-cos 4x)/2#

to have:

#int cos^2xsin^2x dx = 1/8 int (1-cos 4x)dx #

#int cos^2xsin^2x dx = 1/8 int dx - 1/32 int cos 4x d(4x) #

#int cos^2xsin^2x dx = (4x-sin4x)/32 +C#

Jan 17, 2018

#intcos^2xsin^2xdx=1/8x-1/32sin(4x)+"C"#

Explanation:

Given: #intcos^2xsinx^2xdx#

The trick is to use a number of identities:

Use the half angle identities to rewrite the integral:

#color(blue)(sin^2x=1/2(1-cos(2x))#

#color(blue)(cos^2x=1/2(1+cos(2x))#

#=int(1/2(1+cos(2x)))(1/2(1-cos(2x)))dx#

Take out the constants: #(1/2*1/2=1/4)#

#=1/4int(1+cos(2x))(1-cos(2x))dx#

#=1/4int1-cos^2(2x)dx#

Use the identity #color(blue)(sin^2x+cos^2x=1=>1-cos^2x=sin^2x#

#=1/4intsin^2(2x)dx#

Use the identity: #color(blue)(sin^2x=(1-cos(2x))/2#

#=1/4int(1-cos(4x))/2dx#

#=1/4*1/2int1-cos(4x)dx#

#=1/8int1-cos(4x)dx#

Integrate each term:

#=1/8int1dx-intcos(4x)dx#

#=1/8[x-1/4sin(4x)]+"C"#

#=1/8x-1/32sin(4x)+"C"#