What is antiderivative of sin2x*cos2x*cos^24x?

2 Answers | Add Yours

Top Answer

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to use integration to find the antiderivative of `sin2x*cos2x*cos^2 (4x)`

You should use the formula `sin 2x*cos 2x = (sin 2*(2x))/2 = (sin 4x)/2`

Write the transformed function `sin2x*cos2x*cos^2 (4x) = ((sin 4x)*cos^2 (4x))/2` .

Integrating both sides yields:

`int (sin2x*cos2x*cos^2 (4x)) dx= int (((sin 4x)*cos^2 (4x))/2) dx`

You need to write the new function in terms of sin 4x. Use the basic trigonometric formula `sin^2 alpha + cos ^2 alpha = 1` .

`cos^2 (4x) = 1 - sin^2 (4x)`

`sin 4x*cos^2 (4x) = sin 4x*(1 - sin^2 (4x)) = sin 4x - sin^3 (4x).`

`int (((sin 4x)*cos^2 (4x))/2) dx =int ((sin 4x)dx)/2 - int ((sin^3 (4x))dx)/2`

`` `int ((sin^2 (4x)* sin x)dx)/2`

Write `sin x dx = -(cos x)'`

Use integration by parts:

`int udv + int vdu = uv =gt int udv = uv - int vdu`

`u = sin x =gt du = cos x dx`

`` `dv = sin x dx =gt v = - cos x`

`int ((sin^2 (4x)* sin x)dx)/2 = -(sin x*cos x)/2+ (1/2)*int cos^2 x dx`

Use the formula `cos^2 x = (1+cos 2x)/2` `int cos^2 x dx = int dx/2 + (1/2)*int cos 2x dx`

`int cos^2 x dx = x/2 + (sin 2x)/4 + c`

`(1/2)*int cos^2 x dx =x/4 + (sin 2x)/8 + c`

`int ((sin^2 (4x)* sin x)dx)/2 = -(sin x*cos x)/2+ x/4 + (sin 2x)/8 + c`

`int (((sin 4x)*cos^2 (4x))/2) dx = -(cos 4x)/8+(sin x*cos x)/2- x/4- (sin 2x)/8 + c`

The antiderivative of the function is `int (((sin 4x)*cos^2 (4x))/2) dx = -(cos 4x)/8+(sin x*cos x)/2- x/4- (sin 2x)/8 + c.`

beckden's profile pic

beckden | High School Teacher | (Level 1) Educator

Posted on

sin(2A)=2sin(A)cos(A) so substitute  A=2x and we can solve to get


Now use substitution

` u=cos(4x); du = -4sin(4x) dx`

; So `sin(4x)dx=-1/4du`

So our integral is

`int 1/2 sin(4x) cos^2(4x) dx = 1/2 int cos^2(4x) sin(4x) dx = 1/2 int u^2 (-1/4) du`

`=-1/8 (1/3)u^3 + C = -1/24 cos^3(4x) + C`

You can check by taking the derivative.

We’ve answered 317,556 questions. We can answer yours, too.

Ask a question