More integrals

**polymerase** · February 17th 2008, 05:32 PM

$\int \frac{\sec\:x\;\cos\:2x}{\sin\;x+\sec\;x}\;dx$

Thanks

**angel.white** · February 17th 2008, 06:28 PM

Originally Posted by polymerase

$\int \frac{\sec\:x\;\cos\:2x}{\sin\;x+\sec\;x}\;dx$

Thanks

IDK if this is the best way to get the answer, but it's how I found it. Fist, cos2x = cos^2x-sin^2x so
$=\int \frac{sec[(x)cos^2(x)-sin^2(x)]}{sin(x)+sec(x)}dx$

multiply by $\frac{cos(x)}{cos(x)}$
$=\int \frac{cos^2(x)-sin^2(x)}{sin(x)cos(x)+1}dx$

Mess with the denominator:
$=\int \frac{cos^2(x)-sin^2(x)}{\frac 12[2sin(x)cos(x)]+1}dx$

$\begin{array}{|c|} \hline And~you~should~know~that:\\ \hline\\ cos^2(x)=\frac 12[1+cos(2x)]\\\\ sin^2(x)=\frac 12[1-cos(2x)]\\\\ 2sin(x)cos(x)=sin(2x)\\\\\hline\end{array}$

And now we can use these rules:
$=\int \frac{\frac 12(1+cos(2x)-\frac 12(1-cos(2x)}{\frac 12sin(2x)+1}dx$

Simplify
$=\int \frac{\frac 12+\frac 12cos(2x)-\frac 12+\frac 12cos(2x)}{\frac 12sin(2x)+1}dx$

Simplify
$=\int \frac{cos(2x)}{\frac 12sin(2x)+1}dx$

$\begin{array}{|c|} \hline SUBSTITUTION\\ \hline a=sin(2x)\\ da = 2cos(2x)~dx\\ \frac 12~da = cos(2x)~dx\\ \hline\end{array}$

Make substitution
$=\frac 12\int \frac{1}{\frac 12a+1}da$

Distribute the 2 into the denominator
$=\int \frac{1}{a+2}da$

Integrate
$=ln|a+2|+C$

Anti-substitute
$=ln|sin(2x)+2|+C$

I'm pretty sure that's right, at the very least, I like all the steps I used to get there.

**Soroban** · February 17th 2008, 08:13 PM

Hello, polymerase!

This is a shorter version of angel.white's solution . . .

$\int \frac{\sec x\cos2x}{\sin x+\sec x}\,dx$

Multiply by $\frac{2\cos x}{2\cos x}\!:\;\;\int\frac{2\cos2x}{2\sin x\cos x + 2}\,dx \;=\;\int\frac{2\cos2x}{\sin2x + 2}\,dx$

Let $u \:=\:\sin2x + 2\quad\Rightarrow\quad du \:=\:2\cos2x\,dx$

Substitute: . $\int\frac{du}{u} \;=\;\ln|u| + C$

Back-substitute: . $\ln|\sin2x + 2| + C$

**angel.white** · February 17th 2008, 08:56 PM

Originally Posted by Soroban

Hello, polymerase!

This is a shorter version of angel.white's solution . . .

Multiply by $\frac{2\cos x}{2\cos x}\!:\;\;\int\frac{2\cos2x}{2\sin x\cos x + 2}\,dx \;=\;\int\frac{2\cos2x}{\sin2x + 2}\,dx$

Let $u \:=\:\sin2x + 2\quad\Rightarrow\quad du \:=\:2\cos2x\,dx$

Substitute: . $\int\frac{du}{u} \;=\;\ln|u| + C$

Back-substitute: . $\ln|\sin2x + 2| + C$

Well sure, but how are you going to scare anyone with that?!

**Krizalid** · February 18th 2008, 05:07 AM

Fast solutions always are found with practice and of course, experience. (Smart people can get fast solutions too.)

**Soroban** · February 18th 2008, 11:42 AM

I just tripped over the most amazing approach . . .

$\int \frac{\sec x \cos 2x}{\sin x+\sec x}\,dx$

Multiply by $\frac{\cos x}{\cos x}\!:\;\;\int \frac{\cos2x}{\sin x\cos x + 1}\,dx$

Let $u \:=\:\sin x\cos x + 1$

. . Then: . $du \:=\:\left[(\sin x)(-\sin x) + (\cos x)(\cos x)\right]\,dx\;=\;(\cos^2\!x-\sin^2\!x)\,dx \;=\;\cos2x\,dx$

Substitute: . $\int\frac{du}{u} \;=\;\ln|u| + C$

Back-substitute: . $\ln|\sec x\cos x + 1| + C$ . . . ta-DAA!

Is that scary enough?

**Jhevon** · February 18th 2008, 11:44 AM

Originally Posted by Soroban

Is that scary enough?

Terrifying, Soroban, just terrifying!

**angel.white** · February 18th 2008, 01:24 PM

Gonna have to check for it under my bed tonight with a flashlight.

**Krizalid** · February 18th 2008, 02:03 PM

Well anyway to contemplate $u'$ it wasn't necessary to apply the product rule, just note $\sin x\cos x = \frac{{\sin 2x}} {2}$ and the rest follows.

**Jhevon** · February 18th 2008, 02:21 PM

Originally Posted by Krizalid

Well anyway to contemplate $u'$ it wasn't necessary to apply the product rule, just note $\sin x\cos x = \frac{{\sin 2x}} {2}$ and the rest follows.

yup! but that would make it less scary

we have to be quick on one hand, but difficult on the other...

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