with Trig Substitution Integration

[silverdiesel]

I am not too good with trig identities. I can't seem to figure out how to simplify these trig intergrals. I know I can use a triangle to turn the second problem into a trig integral, but once I have the trig integral, I am lost. Any help would be greatly appriciated.

$$\int\tan(x)\sec^3(x)dx$$
$$\int\frac{1}{x^2&\sqrt{16-x^2}}dx$$

 

[Valhalla]

Have you tried a u-substitution on the first one yet? Here is a hint what is the derivative of sec?

 

[silverdiesel]

see, that's what I thought, but there seems to be an extra secant in there. The derivative of sec is sectan. So, u=sec(x), du=sec(x)tan(x)dx. That gives tan(x)sec^2(x)... right?

 

[Tide]

HINT: Let

$$u = \cos x$$

It will save you a lot of work.

 

[Valhalla]

$$u=sec(x)$$ $$du=sec(x) tan(x)dx$$

$$\int u^2du$$

 

[silverdiesel]

brilliant! Thanks Tide. I should have seen that.
$$\int\frac{sin(x)}{cos^4(x)}dx = \frac{1}{3cos^3(x)}$$

 

[silverdiesel]

right, yes I can see that too Valhalla. Thanks so much. Looks like that one was much easier than I made it.

 

[silverdiesel]

Any ideas on the second problem? Using a triangle, I have changed it to:

$$\int\frac{1}{(4sin\Theta)^2(4cos\Theta)}d\Theta$$

 

[silverdiesel]

wait, I think got it, $$u=sin\Theta$$