Please find the value for this integration. What is the answer?

#int_0^(pi/2)sin^3xcosxdx#

2 Answers
Feb 23, 2016

#u = sin(x)#

#du = cos(x)dx#

#int_0^(1)u^3du#

...

Feb 23, 2016

#= 1/4 #

Explanation:

Using the Integral

# intsin^n axcdot cos axcdot dx = 1/[a(n+1)]sin^{n+1} ax +C#
for #(n!=-1)#

Given expression is

# int _0^(pi/2)sin^3 xcdot cos xcdot dx # ,

here #n=3 and a=1#
We obtain #=[ 1/[1(3+1)]sin^{3+1} x +C] _0^(pi/2) #
#=[ 1/4sin^4 x +C] _0^(pi/2) #
Inserting the value of #sin(pi/2)=1 and sin 0=0#
#=[ 1/4xx 1^4 +C] -[ 1/4xx 0^4 +C] #
#= 1/4 #