How Do You Master Trigonometric Integration with Higher Exponentials?

[p4nda]

http://img57.imageshack.us/img57/5647/integraljr9.jpg

Can anybody please help me explain how to do this step-by-step by using the techniques of integration?

Thank you.

 

[VietDao29]

There should be a chapter in your textbook about trigonometric integral. You can take a look at it.
Generally, to integrate:
$$\int \tan ^ {2n}x dx$$, or $$\int \cot ^ {2n}x dx$$, we can do as follow:
Say, we want to integrate:
$$\int \tan ^ 6 x dx = \int ( \tan ^ 6 x + \tan ^ 4 x - \tan ^ 4 x - \tan ^ 2 x + \tan ^ 2 x + 1 - 1) dx$$
$$= \int \tan ^ 4 x (\tan ^ 2 x + 1) dx - \int \tan ^ 2 x (\tan ^ 2 x + 1) dx + \int \sec ^ 2 x dx - \int dx$$
$$= \int \tan ^ 4 x \sec ^ 2 x dx - \int \tan ^ 2 x \sec ^ 2 x dx + \int \sec ^ 2 x dx - \int dx$$
$$= \int \tan ^ 4 x d(\tan x) - \int \tan ^ 2 x d(\tan x) + \int \sec ^ 2 x dx - \int dx$$
$$= \frac{\tan ^ 5 x}{5} - \frac{\tan ^ 3 x}{3} + \tan x - x + C$$.
Ok, can you go from here? :)

 

[p4nda]

For the problem I posted, don't I have to switch "cot" to "cos/sin" (ratio identity) and then use the half-angle/double identities?

 

[VietDao29]

For the problem I posted, don't I have to switch "cot" to "cos/sin" (ratio identity) and then use the half-angle/double identities?

If you change to sin, and cos, and work from there, I think you will be messed up. You'll get some expression quite 'ugly'. You can try to see if it works. :)
Anyway, have you tried the way I suggested? Did you get the answer?
In the example, I used the identity: sec²x = tan²x + 1.
To tackle the problem you asked, you should note that: csc²x = cot²x + 1.

 

[p4nda]

I really appreciate your help, but I'm still kind of lost. I understand how to do the ones with a "6" exponential, but usually the ones with a "4" exponential confuses me. The integrals with a trig. function alone with an "even" exponential (e.g. Tan^[x]) is what gets me. However, I understand the integrals with two trig. functions (e.g. Tan^3[x]Sec^4[x]). For the integral with a single trig. function and a "4" exponential, are you supposed to split it into two (ex. Cot^2[x]Cot^2[x])?

I tried doing my posted problem and I got:
1/3csc^3[x] - cscx + C

 

[VietDao29]

I really appreciate your help, but I'm still kind of lost. I understand how to do the ones with a "6" exponential, but usually the ones with a "4" exponential confuses me. The integrals with a trig. function alone with an "even" exponential (e.g. Tan^[x]) is what gets me. However, I understand the integrals with two trig. functions (e.g. Tan^3[x]Sec^4[x]). For the integral with a single trig. function and a "4" exponential, are you supposed to split it into two (ex. Cot^2[x]Cot^2[x])?

You can add, and then subtract cot²x, and 1 respectively. Like this:
$$\cot ^ 4 x = \cot ^ 4 x + \cot ^ 2 x - \cot ^ 2 x - 1 + 1$$
Then we'll try to factor the expression, and we have:
$$\cot ^ 4 x = \cot ^ 4 x + \cot ^ 2 x - \cot ^ 2 x - 1 + 1 = \cot ^ 2 x (\cot ^ 2 x + 1) - (\cot ^ 2 x + 1) + 1 = \cot ^ 2 x \csc ^ 2 x - \csc x + \fbox{1}$$.
The main aim is to factor it so that csc²x appear, and we can then use the substitution:
$$u = \cot x \Rightarrow du = - \csc ^ 2 x$$
It's the same as my example above. :)

I tried doing my posted problem and I got:
1/3csc^3[x] - cscx + C

Be careful with the signs. And you are forgetting an x there. After factoring, you'll be left with a 1 (see the boxed part above). The answer should be:
something + x + C
Can you go from here? :)