Integral Problem Homework: Attempting Difficult Term

[rjs123]

Homework Statement

trying to integrate this...the second term is the difficult one here.

\theta^2 + 2\theta\sin2\theta + sin^2(2\theta)

The Attempt at a Solution

I attempted the problem and ended up with this but it doesn't seem right

\frac{1}{3}\theta^3-\theta\cos2\theta+ 1/2sin2\theta + \frac{1 - cos4\theta}{2}

 

[Wingeer]

Not quite. Although the two first integrals are correct. Hint: Write u=2theta and use what you know about trigonometric identities.

 

[rjs123]

Not quite. Although the two first integrals are correct. Hint: Write u=2theta and use what you know about trigonometric identities.

i know sin2\theta trig identity is 2sin\theta\cos\theta

u = 2\theta
du = 2d\theta

dv = sin2\theta dv = 2sin\theta\cos\theta


this is where i get stuck

 

[Wingeer]

Know that:
\sin^2(x) = \frac{1-\cos(2x)}{2}.

 

[dextercioby]

Well, what's the derivative of \cos 2\theta equal to ? Then use part integration.

 

[rjs123]

Know that:
\sin^2(x) = \frac{1-\cos(2x)}{2}.

\theta^2 + 2\theta\sin2\theta + sin^2(2\theta)

after integrating...

\frac{1}{3}\theta^3-\theta\cos2\theta+ 1/2sin2\theta + \frac{1}{2}\theta - \frac{1}{8}sin4\theta

i forgot to integrate the last part...but this still doesn't seem correct

 

[Wingeer]

But it is correct. Up to a constant, of course.

 

[rjs123]

But it is correct. Up to a constant, of course.

I'm attempting to solve this area problem


1/2\int_{0}^{\pi }(\theta + sin2\theta)^2 d\theta}


The area found by my calculator comes out to be 4.93...but by hand I get 4.38


The original polar equation: r = \theta + sin(2\theta) from 0 to pi.

I think it may by the use of my input into the calculator and not the work done by hand...i'll doublecheck.

 

[SammyS]

<br /> \frac{1}{3}\theta^3-\theta\cos2\theta+ 1/2\sin2\theta + \frac{1}{2}\theta - \frac{1}{8}\sin4\theta<br />

Add a constant & it looks good to me. Check it by taking the derivative.

 

[rjs123]

<br /> \frac{1}{3}\theta^3-\theta\cos2\theta+ 1/2\sin2\theta + \frac{1}{2}\theta - \frac{1}{8}\sin4\theta<br />

Add a constant & it looks good to me. Check it by taking the derivative.

yep...i just concluded guys that I was inputting the equation wrong into my calculator...4.38 is the right answer and i was doing it right by hand all along...what a relief.

I have one more question though...

how do I find the angle at which the graph is at x = -2 ?