How do you find the integral of #x(sin^2(ax))#?

Question

Emilia Aguilar · Accepted Answer

#(2a^2x^2-2axsin(2ax)-cos(2ax))/(8a^2)+C#

#I=intxsin^2(ax)dx#

First, let #t=ax#. This implies that #dt=adx#. Note that this means that #x=t/a# and #dx=(dt)/a#.

#I=intt/asin^2(t)dt/a=1/a^2inttsin^2(t)dt#

This will become easier if we rewrite #sin^2(t)# to make it easier to manipulate, a common trick in integrals of this sort. We will use one version of the cosine double angle formula:

#cos(2t)=1-2sin^2(t)" "=>" "sin^2(t)=1/2(1-cos(2t))#

This equivalency is used to transform the integral into:

#I=1/a^2intt/2(1-cos(2t))dt#

#I=1/(2a^2)int(t-tcos(2t))dt#

Dividing the integral:

#I=1/(2a^2)inttdt-1/(2a^2)inttcos(2t)dt#

The power rule for integration makes it simple to complete the first integral:

#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 #s=2t#, implying that #ds=2dt#. Thus, #dt=(ds)/2# and #t=s/2#.

#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 #intudv=uv-intvdu#. We will let:

#{(u=s" "=>" "du=ds),(dv=cos(s)ds" "=>" "v=sin(s)):}#

Using the following in the formula:

#I=t^2/(4a^2)-1/(8a^2)[ssin(s)-intsin(s)ds]#

Since #intsin(s)ds=-cos(s)#:

#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 #s=2t#:

#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 #t=ax#:

#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)#

Should we seek a shared denominator:

#I=(2a^2x^2-2axsin(2ax)-cos(2ax))/(8a^2)+C#

Aurora Alvarado · Answer

undefined To find the integral of x(sin^2(ax)), we can use integration by parts. Let u = x and dv = sin^2(ax) dx. Then, du = dx and v = (1/2)(x - (1/(2a))sin(2ax)). Applying the integration by parts formula, we have:

∫ x(sin^2(ax)) dx = (1/2)x^2 - (1/(2a)) ∫ x sin(2ax) dx

To evaluate the remaining integral, we use integration by parts again. Let u = x and dv = sin(2ax) dx. Then, du = dx and v = -(1/(2a))cos(2ax). Applying the integration by parts formula, we have:

∫ x sin(2ax) dx = -(1/(4a))x cos(2ax) + (1/(4a^2)) ∫ cos(2ax) dx

= -(1/(4a))x cos(2ax) + (1/(8a^3))sin(2ax) + C

Substituting this result back into the original equation, we have:

∫ x(sin^2(ax)) dx = (1/2)x^2 - (1/(2a))(-(1/(4a))x cos(2ax) + (1/(8a^3))sin(2ax)) + C

= (1/2)x^2 + (1/(8a^2))sin(2ax) - (1/(8a^3))x cos(2ax) + C