Trigonometric Substitution for Integrating Radical Expressions

[silicon_hobo]

Homework Statement

Hey, it's me again. This method is giving me some trouble. This is the first problem: $$\int^3_0\ x^2\sqrt{9-x^2} \ dx$$

The second problem is:
$$\int\frac{dx}{\sqrt{2x^2+2x+5}}$$. How do I use a trig. substitution to start on this one?

Homework Equations

The Attempt at a Solution

http://www.mcp-server.com/~lush/shillmud/int2.4a.JPG
I know I need to apply an identity here and then maybe integrate by parts. Also, what's the proper way to transform the limits of integration in this type of substitution? Thank you for your input.

 

[rocomath]

Evaluate for your new limits upon substitution.

$$x=3\sin\theta$$

$$\int_0^3\rightarrow\int_0^\frac{\pi}{2}$$

$$81\int_0^\frac{\pi}{2}\sin^{2}\theta\cos^{2}\theta d\theta$$

$$81\int_0^\frac{\pi}{2}\sin^{2}\theta\cos\theta\cos\theta d\theta$$

$$u=\cos\theta$$
$$du=-\sin\theta d\theta$$

$$dV=\sin^{2}\theta\cos\theta d\theta$$
$$V=\frac 1 3\sin^{3}\theta$$

Use parts and you will notice it is a recursive ... bring your original Integral to the left and all you have to evaluate is $$\int\sin^{2}\theta d\theta$$ which can be simplified using a trig identity $$\sin^{2}\theta=\frac 1 2 (1-\cos{2x})$$

 

[rocomath]

Ok done typing.

For your 2nd problem, complete the square and use a Trig sub! Make sure that your leading term is positive and one.

 

[silicon_hobo]

Okay, I think we agree on the first one:
http://www.mcp-server.com/~lush/shillmud/int2.4a2.JPG
But how do I get rid of that pesky $$d\theta$$?

This is what I've got so far for #2. I'm not sure if I've applied the identity correctly:
http://www.mcp-server.com/~lush/shillmud/int2.4b.JPG

 

[rocomath]

Ah very nice alternative to what I suggested, but don't you love that though ... works both ways! Also, keep in mind what I did ... b/c it becomes very useful to notice the chain rule. What do you mean get rid of d-theta? You evaluated for your new limits, so you don't need to get rid of it.

$$\frac{81}{8}\int_0^\frac{\pi}{2}(1-\cos{4\theta})d\theta$$

#2, you made a mistake when you factored out the 2.

$$\int\frac{dx}{\sqrt{2x^2+2x+5}}$$

Evaluating only the radican ...

$$2x^2+2x+5 \rightarrow 2\left(x^2+x+\frac 5 2\right)$$

You applied everything correctly, now go back and just fix the factoring error.