Integral of cosine squared x?

**grgrsanjay** · May 6th 2010, 06:54 AM

use integration by parts and let u=cosx and dv=cosxdx
du=-sinx v=sinx
Int(udv)=uv-Int(vdu) so uv=cosx(sinx) and vdu=sinx(-sinx)
so we have:

Int(cos^2(x)=(cosx)(sinx)+Int(sin^2x)
(the (-) became + because of the -sinx, so we add Int(vdu))
Now it looks not better because we have sin^2x instead of cos^2x,
but sin^2x=1-cos^2x since sin^2x+cos^2x=1
So we have

Int(cos^2x)=cosxsinx+Int(1-cos^2x)
=cosxsinx+Int(1)-Int(cos^2x)
So now add the -Int(cos^2x) on the RHS to the one on the LHS
2Int(cos^2x)=cosxsinx+x
so Int(cos^2x)=1/2[cosxsinsx+x] and now add the constant!
final answer

Integral of cos^2x=(1/2)(cosx sinx + x)+C = x/2 + (1/4)sin 2x + C
(because sin x cos x = (1/2)sin 2x)

right??

**amul28** · May 6th 2010, 07:45 AM

the answer is correct...
but is there no other way...
it is like round up.

**HallsofIvy** · May 6th 2010, 07:52 AM

Simpler to use the trig identity $cos^2(x)= \frac{1}{2}(1+ 2cos(x))$ so that $\int cos^2(x)dx= \frac{1}{2}\int dx+ \int cos(x)dx= \frac{1}{2}x+ sin(x)+ C$ .

**amul28** · May 6th 2010, 07:57 AM

Originally Posted by HallsofIvy

Simpler to use the trig identity $cos^2(x)= \frac{1}{2}(1+ 2cos(x))$ so that $\int cos^2(x)dx= \frac{1}{2}\int dx+ \int cos(x)dx= \frac{1}{2}x+ sin(x)+ C$ .

i dont think ur answer is equal to cos^2(x)

**grgrsanjay** · May 6th 2010, 08:30 AM

Originally Posted by HallsofIvy

Simpler to use the trig identity $cos^2(x)= \frac{1}{2}(1+ 2cos(x))$ so that $\int cos^2(x)dx= \frac{1}{2}\int dx+ \int cos(x)dx= \frac{1}{2}x+ sin(x)+ C$ .

how did you get this identity??

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dint ever heard of it

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