# Easy looks integral. Help

1. Feb 21, 2007

### alba_ei

Hello I have this integral and I can't solve it

$$\int\csc^6 x dx$$

Well i first start with trig identnity $$\csc x = \frac{1}{sen x}$$ so my first attempt to slove it looks like this

$$\int\frac{dx}{(\sin^2 x)^3}$$

then i subsituted with $$\sin^2 x = \frac{1 - \cos 2x}{2}$$

and i get

$$8\int\frac{dx}{(1-cos 2x)^3}$$ and i saw that i was going to

nowhere so i try another way

i took the function and use the trig identity to get this

$$\int\csc^6 x dx$$ =$$\int \frac{\sec^6 x}{\tan^6 x} dx$$

$$u = \tan x$$ $$du = \sec^2 x$$

then

$$\int\frac{du^3}{tan^6 x}$$

and i get stucked again. any ideas?

Last edited: Feb 21, 2007
2. Feb 22, 2007

### neutrino

That looks most promising.

If $$u = \tan{x}$$, then $$du = \sec^2{x}dx$$ (Don't forget the dx)

Now, rewrite the integral as

$$\int{\frac{\sec^4{x}\sec^2{x}dx}{\tan^6{x}}}$$

Now, think of an identity that relates $$\sec{x}$$ and $$\tan{x}$$.

Last edited: Feb 22, 2007
3. Feb 22, 2007

### dextercioby

Make $\tan x =t$ and then u'll get 3 simple integrals.

4. Feb 22, 2007

### VietDao29

Here's the trick, if the power of csc function is even, you can use the u-substitution u = cot(x).
And if the power of sec function is even, you can use the u-substitution u = tan(x)
So, in your problem, the power of csc function is even, we use the u-substitution: u = cot(x)
du = -csc2(x) dx, so your integral will become:
$$\int \csc ^ 6 (x) dx = \int ( \csc ^ 4 (x) \times \csc ^ 2 (x) ) dx = \int {( \csc ^ 2 (x) )} ^ 2 \times \csc ^ 2 (x) dx = \int {\left( 1 + \cot ^ 2 (x) \right)} ^ 2 \times \csc ^ 2 (x) dx$$
$$= -\int {(1 + u ^ 2)} ^ 2 du = ...$$
Can you go from here? :)