How do you find the integral of #x(sin^2(ax))#?
1 Answer
Explanation:
#I=intxsin^2(ax)dx#
First, let
#I=intt/asin^2(t)dt/a=1/a^2inttsin^2(t)dt#
This will become easier if we rewrite
#cos(2t)=1-2sin^2(t)" "=>" "sin^2(t)=1/2(1-cos(2t))#
Using this equivalency, the integral becomes:
#I=1/a^2intt/2(1-cos(2t))dt#
#I=1/(2a^2)int(t-tcos(2t))dt#
Splitting up the integral:
#I=1/(2a^2)inttdt-1/(2a^2)inttcos(2t)dt#
The first integral can be performed easily using the power rule for integration:
#I=1/(2a^2)(t^2/2)-1/(2a^2)inttcos(2t)dt#
#I=t^2/(4a^2)-1/(2a^2)inttcos(2t)dt#
This is another technically unnecessary substitution, but it clears up the following steps. Let
#I=t^2/(4a^2)-1/(2a^2)ints/2cos(s)(ds)/2#
#I=t^2/(4a^2)-1/(8a^2)intscos(s)ds#
For this integral, we will use integration by parts, which takes the form
#{(u=s" "=>" "du=ds),(dv=cos(s)ds" "=>" "v=sin(s)):}#
Plugging these into the formula:
#I=t^2/(4a^2)-1/(8a^2)[ssin(s)-intsin(s)ds]#
Since
#I=t^2/(4a^2)-1/(8a^2)(ssin(s)+cos(s))#
#I=t^2/(4a^2)-(ssin(s))/(8a^2)-cos(s)/(8a^2)#
Back-substituting with
#I=t^2/(4a^2)-(2tsin(2t))/(8a^2)-cos(2t)/(8a^2)#
#I=t^2/(4a^2)-(tsin(2t))/(4a^2)-cos(2t)/(8a^2)#
Back-substituting with
#I=(ax)^2/(4a^2)-(axsin(2ax))/(4a^2)-cos(2ax)/(8a^2)#
#I=(a^2x^2)/(4a^2)-(xsin(2ax))/(4a)-cos(2ax)/(8a^2)#
#I=(x^2)/4-(xsin(2ax))/(4a)-cos(2ax)/(8a^2)#
If we want a common denominator:
#I=(2a^2x^2-2axsin(2ax)-cos(2ax))/(8a^2)+C#