integral with cotx

**chocolatelover** · September 7th 2007, 01:23 PM

Hi,

Could someone please give me a hint on how to do this problem? I know that I have to use a formula to do it, but it looks like I need to simplify it.

Int. cotx/sq. 1+2sinx

I have these formulas:

int. cotudu=ln|secu|+c

int. cot^udu=-1/n-1cot^n-1u-int. cot^n-2udu

Thank you

**red_dog** · September 7th 2007, 01:35 PM

$\displaystyle\int\frac{\cot x}{\sqrt{1+2\sin x}}dx=\int\frac{\cos x}{\sin x\sqrt{1+2\sin x}}dx$
Substitute $\sin x=t\Rightarrow\cos xdx=dt\Rightarrow\int\displaystyle\frac{1}{t\sqrt{ 1+2t}}dt$ .
Substitute $\sqrt{1+2t}=u\Rightarrow 2t=u^2-1\Rightarrow dt=udu$
Then $\displaystyle\int\frac{u}{\frac{u^2-1}{2}\cdot u}du=2\int\frac{1}{u^2-1}du=\ln \left|\frac{u-1}{u+1}\right|+C$
Now back substitute.

**Soroban** · September 7th 2007, 01:54 PM

Hello, chocolatelover!

I found a method . . . but it took two substitutions.
I'm certain someone will streamline it for us.

$\int \frac{\cot x}{\sqrt{1+2\sin x}}\,dx$

Let $u \,=\,\sin x\quad\Rightarrow\quad x \,=\,\arcsin u\quad\Rightarrow\quad dx \,=\,\frac{du}{\sqrt{1-u^2}}$
. . and: . $\cot x \,=\,\frac{\sqrt{1-u^2}}{u}$

Substitute: . $\int\frac{\frac{\sqrt{1-u^2}}{u}}{\sqrt{1+2u}}\,\frac{du}{\sqrt{1-u^2}} \;= \;\int\frac{du}{u\sqrt{1+2u}}$

Let $v \,=\,\sqrt{1+2u}\quad\Rightarrow\quad u \,=\,\frac{v^2-1}{2}\quad\Rightarrow\quad du \,=\,v\,dv$

Substitute: . $\int\frac{v\,dv}{\left(\frac{v^2-1}{2}\right)\cdot v} \;=\;2\int\frac{dv}{v^2-1} \;= \;\ln\left|\frac{v-1}{v+1}\right| + C$

Back-substitute: . $\ln\left|\frac{\sqrt{1+2u} - 1}{\sqrt{1+2u} + 1}\right| + C$

Back-substitute: . $\boxed{\ln\left|\frac{\sqrt{1+2\sin x} - 1}{\sqrt{1+2\sin x} + 1}\right| + C}$

Ha! . . . red_dog beat me to it!

**Krizalid** · September 7th 2007, 02:06 PM

Mine just take one.

Set

$u=\sqrt{1+2\sin x}\implies du=\frac{\cos x}{\sqrt{1+2\sin x}}\,dx$ , the intregral becomes to

$\int\frac{\cot x}{\sqrt{1+2\sin x}}\,dx=2\int\frac1{u^2-1}\,du$

Cheers,
K.

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