# Evaluating the Integrals

• December 9th 2009, 05:16 PM
Gilvanildo
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?
• December 9th 2009, 05:30 PM
skeeter
$\int \csc^4{x}\cot^6{x} \, dx$

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

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

$u = \cot{x}$

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

$-\int (u^8 + u^6) \, du$
• December 9th 2009, 05:38 PM
Prove It
Quote:

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 $x + 1 = 2\sec{\theta}$.
• December 9th 2009, 05:51 PM
Soroban
Hello, Gilvanildo!

Quote:

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

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 . . .

Quote:

$(b)\;\int\frac{x\,dx}{\sqrt{x^2+2x-3}}$
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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