How do you evaluate the definite integral #int sin^3xcos^4xdx# from #[0, pi/2]#?

Question

6224 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-06T17:46:37

#2/35#

Turn to the form #int(sinx(cos^nx - cos^mx))dx# and then integrate using the substitution #u = cosx#.

#=int_0^(pi/2) sinx(sin^2x(cos^4x))dx#

#=int_0^(pi/2) sinx(1 - cos^2x)cos^4xdx#

#=int_0^(pi/2) sinx(cos^4x - cos^6x)dx#

Let #u = cosx#. Then #du = -sinxdx#, and #dx= -(du)/sinx#.

#=int_0^(pi/2) sinx(u^4 - u^6) * -(du)/sinx#

#=-[1/5u^5 - 1/7u^7]_0^(pi/2)#

#=-[1/5cos^5x - 1/7cos^7x]_0^(pi/2)#

Evaluate using the 2nd fundamental theorem of calculus.

#=-(1/5cos^5(pi/2) - 1/7cos^7(pi/2) - (1/5cos^5(0) - 1/7cos^7(0)))#

#=-(0 - 1/5 + 1/7)#

#=2/35#

Hopefully this helps!