Indefinite integral of sin^4xdx

[Captain Zappo]

Hi everyone. I'm having some trouble evaluating the following integral

<br /> \int{sin^4xdx}<br />

First let me start off by showing what I did.

<br /> = \int{(sin^2x)(sin^2x)dx}<br />

<br /> =\int{[\frac{1}{2}-\frac{1}{2}cos(2x)] \ [\frac{1}{2}-\frac{1}{2}cos(2x)]dx<br />

<br /> =\int{(\frac{1}{4}-cos(2x)+\frac{1}{4}cos^2(2x))dx<br />

<br /> =\frac{1}{4}x-\frac{1}{2}sin(2x)+\frac{1}{4}\int{(\frac{1}{2}+\frac{1}{2}cos(4x))dx<br />

<br /> =\frac{1}{4}x-\frac{1}{2}sin(2x)+\frac{1}{8}x+\frac{1}{8}\int{cos(4x)}dx<br />

<br /> =\frac{1}{4}x-\frac{1}{2}sin(2x)+\frac{1}{8}x+\frac{1}{32}sin(4x)+C<br />


I'm not sure if I have the right answer or not, so could someone please check and see if I did anything illegal. If so, can you please correct me, or even show me a different method entirely.

Thanks,
-Zach

 

[quasar987]

<br /> =\int{[\frac{1}{2}-\frac{1}{2}cos(2x)] \ [\frac{1}{2}-\frac{1}{2}cos(2x)]dx<br />

<br /> =\int{(\frac{1}{4}-\frac{1}{2}cos(2x)+\frac{1}{4}cos^2(2x))dx<br />

no?

 

[Captain Zappo]

<br /> =\int{[\frac{1}{2}-\frac{1}{2}cos(2x)] \ [\frac{1}{2}-\frac{1}{2}cos(2x)]dx<br />

<br /> =\int{(\frac{1}{4}-\frac{1}{2}cos(2x)+\frac{1}{4}cos^2(2x))dx<br />

no?

Thank you!

 

[c-murda]

why is the second cos squared?


what is next step after that? I am stuck here too...
a lil hint please

 

[cakesama]

expand the expression in the integral, then you get the last term being cos sqaured

 

[c-murda]

what about after that...

 

[cakesama]

so you expand to get this

=\int{(\frac{1}{4}-\frac{1}{2}cos(2x)+\frac{1}{4}cos^2(2x))dx

then its an integral of 3 terms, so you integrate the 3 terms separately
did that help?

 

[Dick]

what about after that...

There's a trig formula for cos^2(2x) in terms of cos(4x). Just look at what Captain Zappo tried to do, in spite of fluffing a coefficient.

 

[c-murda]

∫sin^4(x)

[sin^2(x)]^2

[(1 -cos(2x)/2)]^2

1/4[1 + cos^2(2x) - 2cos(2x)]

1/4[1 - 2cos(2x) + (1 + cos(4x)/2 ]

1/4) - (1/2)cos(2x) + (1/8) + (1/8)cos(4x)

(3/8) - (1/2)cos(2x) + (1/8)cos(4x)

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

∫sin^4(x)dx = (3/8)x - (1/2)sin(2x)*(1/2) + (1/8)sin(4x)*(1/4) + c

∫sin^4(x)dx = (3/8)x - (1/4)sin(2x) + (1/32)sin(4x) + c

is this right?

 

[Dick]

Looks fine to me.

 

[rootX]

http://integrals.wolfram.com/index.jsp?expr=(sin(x))^4&random=false

You should use this (for other questions)

 

[m_s_a]

∫sin^4(x)dx = (3/8)x - (1/4)sin(2x) + (1/32)sin(4x) + c
correct