Question

Intergal of 8sin^4(x)dx

Answer 1

`\int\limits8sin^4(x)dx =`

Take the constant out
`\int\limits a*f(x)dx=a* \int\limitsf(x)dx`


`= 8 \int\limits sin^4(x)dx`


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


`= 8( -(cos(x)sin^3(x))/(4)+ (3)/(4) \int\limits sin^2(x)dx)`

Use the following identify :
`sin^2(x) = (1-cos(2x))/(2)`


`= 8(- (cos(x)sin^3(x))/(4) + (3)/(4) (1)/(2)( \int\limits 1 - cos(2x)dx)`


`\int\limits 1dx = x`


`\int\limits cos(2x)dx = (1)/(2) sin(2x)`

Apply the sum rule :


`=8(- (cos(x)sin^3(x))/(4) + (3)/(4) (1)/(2) (x- (1)/(2) sin(2x))`

Simplify :


`8( (3)/(8) (x- (1)/(2) sin(2x)) - (1)/(4) sin^3(x)cos(x))`

Therefore add a constant to the solution:


`= 8 ( (3)/(8) (x- (1)/(2) sin(2x)) - (1)/(4) sin^3(x)cos(x))+C`

hope this helps!