Question

How do you find the integral of `((sin(x/5))^2)*((cos(x/5))^3)`?

Answer 1

`=5/48(6sin(x/5)-5sin(3/5x)-sinx)+C`

Explanation:

Use #sin 2A = 2 sin A cos A, sin^2 A =(1-cos 2A)/2 and cos (A + B ) +

cos ( A - B )=2 cos A cos B#.

The integrand is

`(sin(x/5)cos(x/5))^2cos(x/5)`

`=1/4sin^2(2/5x)cos(x/5)`

`=1/8(1-cos(4/5x))cos(x/5)`

`=1/8(cos(x/5)-1/2(cosx + cos(3/5x)))`.

So, the integral is

`1/16[2int cos(x/5) dx- int cos x dx - int cos(3/5x) dx]`

`1/16[10 sin(x/5)-sin x-5/3sin(3/5x)] +C`

`=5/48(6sin(x/5)-5sin(3/5x)-sinx)+C`

Answer 2

`=5/3sin^3(x/5)-sin^5(x/5)+C`

Explanation:

Let `u=sin(x/5)`
`" "`
`color(blue)(du=1/5cos(x/5)dxrArr5du=cos(x/5)dx`
`" "`
Let us start computing the integral by substituting `" "u" and "du`
`" "`
then , using the trigonometric identity#" "color(purple) #" "# (cos^2alpha=1-sin^2alpha)#.
`" "`
`int(sin(x/5))^2xx(cos(x/5))^3dx`
`" "`
`=int(sin(x/5))^2xx(cos(x/5))^2xxcos(x/5)dx`
`" "`
`=int(sin(x/5))^2xx(cos(x/5))^2xxcolor(blue)(5du)`
`" "`
`=intu^2xx(1-sin^2(x/5))xx5du`
`" "`
`=intu^2xx(1-u^2)xx5du`
`" "`
`=5intu^2-u^4du`
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`=5(intu^2du-intu^4)du`
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`=5intu^2du-5intu^4du`
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`=5/3u^3-5/5u^5+C" ,"C " "`is a constant.
`" "`
`=5/3u^3-u^5+C`
`" "`
`=5/3sin^3(x/5)-sin^5(x/5)+C`