# More integrals

• February 17th 2008, 06:32 PM
polymerase
More integrals
$\int \frac{\sec\:x\;\cos\:2x}{\sin\;x+\sec\;x}\;dx$

Thanks
• February 17th 2008, 07:28 PM
angel.white
Quote:

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.
• February 17th 2008, 09:13 PM
Soroban
Hello, polymerase!

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

Quote:

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

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: . $\ln|\sin2x + 2| + C$

• February 17th 2008, 09:56 PM
angel.white
Quote:

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?! (Nerd)
• February 18th 2008, 06:07 AM
Krizalid
Fast solutions always are found with practice and of course, experience. (Smart people can get fast solutions too.)
• February 18th 2008, 12:42 PM
Soroban
I just tripped over the most amazing approach . . .

Quote:

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

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?

• February 18th 2008, 12:44 PM
Jhevon
Quote:

Originally Posted by Soroban

Is that scary enough?

Terrifying, Soroban, just terrifying!
• February 18th 2008, 02:24 PM
angel.white
Gonna have to check for it under my bed tonight with a flashlight. (Sweating)
• February 18th 2008, 03:03 PM
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.
• February 18th 2008, 03:21 PM
Jhevon
Quote:

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 (Nerd)

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