Math Help - Evaluating the Integrals

1. Evaluating the Integrals

I've got these two integral problems, and I'm stuck with these.
Can anybody give me some hints for these two questions?

2. $\int \csc^4{x}\cot^6{x} \, dx$

$\int \csc^2{x}(\cot^2{x}+1)\cot^6{x} \, dx$

$-\int (\cot^8{x} + \cot^6{x})(-\csc^2{x}) \, dx$

$u = \cot{x}$

$du = -\csc^2{x} \, dx$

$-\int (u^8 + u^6) \, du$

3. Originally Posted by Gilvanildo
I've got these two integral problems, and I'm stuck with these.
Can anybody give me some hints for these two questions?
For b), try completing the square in the denominator, then using the substitution $x + 1 = 2\sec{\theta}$.

4. Hello, Gilvanildo!

$(a)\;\int \csc^4\!x\cot^6\!x\,dx$

We have: . $\csc^4\!x\cot^6\!x \;=\;\csc^2\!x\cot^6\!x\,(\csc^2\!x\,dx)$

. . . . . . . . . . . . . . . $= \;(\cot^2\!x+1)\cot^6\!x\,(\csc^2\!x\,dx)$

. . . . . . . . . . . . . . . $= \;(\cot^8\!x + \cot^6\!x)\,(\csc^2\!x\,dx)$

Then we have: . $\int(\cot^8\!x + \cot^6\!x)\,(\csc^2\!x\,dx)$

Let: . $u \,=\,\cot x \quad\Rightarrow\quad dx \,=\,-\csc^2\!x\,dx \quad\Rightarrow\quad \csc^2\!x\,dx \,=\,-du$

. . and substitute . . .

$(b)\;\int\frac{x\,dx}{\sqrt{x^2+2x-3}}$
Under the radical, complete the square: . $\sqrt{x^2+2x-3} \:=\:\sqrt{(x+1)^2 - 4}$

Then we have: . $\int\frac{x\,dx}{\sqrt{(x+1)^2-4}}$

Let: . $x + 1 \:=\:2\sec\theta \quad\Rightarrow \quad dx \:=\:2\sec\theta\tan\theta\,d\theta$

Also: . $x \:=\:2\sec\theta - 1 \quad\hdots\quad \sqrt{(x+1)^2-4} \:=\:2\tan\theta$

Substitute: . $\int\frac{(2\sec\theta-1)(2\sec\theta\tan\theta\,d\theta)}{2\tan\theta} \;=\; \int(2\sec^2\!\theta - \sec\theta)\,d\theta$

Got it?

Edit: Too slow again . . . *blush*
.