Question

How do you integrate `\int _ { 0} ^ { \pi / 2} \cos ^ { 6} x d x`?

Answer 1

Compute the indefinite integral by recursively using the reduction formula,
`intcos^n(x)dx= 1/(n)cos^(n-1)(x)sin(x)+(n-1)/nintcos^(n-2)(x)dx`
, then evaluate at the limits.

Explanation:

Use `intcos^n(x)dx= 1/(n)cos^(n-1)(x)sin(x)+(n-1)/nintcos^(n-2)(x)dx` where n = 6:

`intcos^6(x)dx= 1/(6)cos^5(x)sin(x)+5/6intcos^4(x)dx`

Use `intcos^n(x)dx= 1/(n)cos^(n-1)(x)sin(x)+(n-1)/nintcos^(n-2)(x)dx` where n = 4:

`intcos^6(x)dx= 1/(6)cos^5(x)sin(x)+5/6[1/4cos^3(x)sin(x)+3/4intcos^2(x)dx]`

Use `intcos^n(x)dx= 1/(n)cos^(n-1)(x)sin(x)+(n-1)/nintcos^(n-2)(x)dx` where n = 2:

`intcos^6(x)dx= 1/(6)cos^5(x)sin(x)+5/6[1/4cos^3(x)sin(x)+3/4{1/2cos(x)sin(x)+1/2x}]+C`

Simplify:

`intcos^6(x)dx= 1/(6)cos^5(x)sin(x)+5/24cos^3(x)sin(x)+15/24{1/2cos(x)sin(x)+1/2x}+C`

`intcos^6(x)dx= 1/(6)cos^5(x)sin(x)+5/24cos^3(x)sin(x)+15/48cos(x)sin(x)+15/48x+C`

Evaluate at the limits:

`int_0^(pi/2)cos^6(x)dx= [1/(6)cos^5(x)sin(x)+5/24cos^3(x)sin(x)+15/48cos(x)sin(x)+15/48x]_0^(pi/2)`

`int_0^(pi/2)cos^6(x)dx= (5pi)/32`