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#