Question

Question #642d1

Answer 1

`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`

Answer 2

`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"`