integration by parts for 3(sin(x))^4(cos(x))^2dx

moy1989

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Oct 11, 2007
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Hey folks, I have been trying to integrate the following:

3(sin(x))^4(cos(x))^2dx

but I can't get it, twice I got the same wrong answer and then a different wrong answer. I used integration by parts as a technique although, now, it seems like a bad technique for this problem. I've tried using pythagorean identities and half-angle formulas, but there is something I haven't been able to see. Can you guys help? I'd really appreciate it.
 
moy1989 said:
Hey folks, I have been trying to integrate the following:

3(sin(x))^4(cos(x))^2dx

= 3/4 * [sin(2x)]^2 * [{1 - cos(2x)}/2] dx

Now continue.....
 
Hello, moy1989!

Use these identities:

. . sin2θ=1cos2θ2\displaystyle \sin^2\theta \:=\:\frac{1\,-\,\cos2\theta}{2}

. . cos2θ=1+cos2θ2\displaystyle \cos^2\theta \:=\:\frac{1\,+\,\cos2\theta}{2}


\(\displaystyle \L 3\int \sin^4x\,\cos^2x\,dx\)


sin4xcos2x  =  (sin2x)2cos2x  =  (1cos2x2)2(1+cos2x2)  =  18(1cos2xcos22x+cos32x)\displaystyle \sin^4x\cdot\cos^2x\;=\;\left(\sin^2x\right)^2\cos^2x \;=\;\left(\frac{1\,-\,\cos2x}{2}\right)^2\left(\frac{1\,+\,\cos2x}{2}\right)\;=\;\frac{1}{8}\,\left(1\,-\,\cos2x\,-\,\cos^22x\,+\,\cos^32x\right)

The third term is: cos22x=1+cos4x2=12+12cos4x\displaystyle \:\cos^22x \:=\:\frac{1\,+\,\cos4x}{2}\:=\:\frac{1}{2}\,+\,\frac{1}{2}\cdot\cos4x

The last term is: \(\displaystyle \:\cos^32x \:=\:\cos^22x\cdot\cos2 x \:=\:(1\,-\,\sin^22x)\cos2x \:=\:\cos2x\,-\,\sin^22x\cdot\cos2x\)


Hence, we have: 18[1cos2x(12+12cos4x)+(cos2xsin22xcos2x)]\displaystyle \:\frac{1}{8}\,\left[1\,-\,\cos2x\,-\,\left(\frac{1}{2}\,+\,\frac{1}{2}\cdot\cos4x\right)\,+\,\left(\cos2x\,-\,\sin^22x\cdot\cos2x\right)\right]

. . which simplifies to: 18(1212cos4xsin22xcos2x)\displaystyle \:\frac{1}{8}\,\left(\frac{1}{2}\,-\,\frac{1}{2}\cdot\cos4x\,-\,\sin^22x\cdot\cos2x\right)


The integral becomes: \(\displaystyle \L\:\frac{3}{8}\int \left(\frac{1}{2}\,-\,\frac{1}{2}\cdot\cos4x\,-\,\sin^22x\cdot\cos2x\right)\,dx\)

. . Got it?

 
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