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

2 Answers
Nov 27, 2016

=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

Nov 27, 2016

=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
" "
=5(intu^2du-intu^4)du
" "
=5intu^2du-5intu^4du
" "
=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