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

[moy1989]

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.

 

[Deleted member 4993]

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.....

 

[soroban]

Hello, moy1989!

Use these identities:

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

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

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

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

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

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


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

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


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?

 

[moy1989]

Thank you soroban.