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]

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

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

 

[rjs123]

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

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 ?