Integral of the power products following integral:#int_0^(pi/2)sen^2(θ/3)dθ# ?

1 Answer
Apr 5, 2018

#I=pi/4-(3sqrt3)/8#

Explanation:

We know that,

#color(red)((1)sin^2A=(1-cos2A)/2#

#(2)# If #color(blue)(intf(x)dx=F(x) ,then, intf(ax+b)dx=1/aF(ax+b)+c#

Here,

#I=int_0^(pi/2)sin^2(theta/3)d theta#

#=int_0^(pi/2)color(red)(((1-cos((2theta)/3))/2))d theta...toApply(1)#

#=1/2int_0^(pi/2)1d theta-1/2int_0^(pi/2)cos((2theta)/3)d theta#

#=1/2[theta]_0^(pi/2)-1/2[color(blue)((sin((2theta)/3))/(2/3))]_0^(pi/2)...toApply (2)#

#=1/2[pi/2-0]-3/4[sin(2/3(pi/2))-sin0]#

#=pi/4-3/4[sin(pi/3)-0]#

#=pi/4-3/4*sqrt3/2#

#=pi/4-(3sqrt3)/8~~0*135#
Note:
We can take #(2theta)/3=u or apply(2)#
i.e.#theta=3/2u=>d theta=3/2du#
Also, #theta=0=>u=0and theta=pi/2=>u=2/3(pi/2)=pi/3#
and so on...