Evaluating the Integrals

**Gilvanildo** · December 9th 2009, 05:16 PM

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

**skeeter** · December 9th 2009, 05:30 PM

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

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

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

$u = \cot{x}$

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

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

**Prove It** · December 9th 2009, 05:38 PM

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?
Thanks for answering.

For b), try completing the square in the denominator, then using the substitution $x + 1 = 2\sec{\theta}$ .

**Soroban** · December 9th 2009, 05:51 PM

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*
.

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