1. ## More integrals

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

Thanks

2. Originally Posted by polymerase
$\displaystyle \int \frac{\sec\:x\;\cos\:2x}{\sin\;x+\sec\;x}\;dx$

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
$\displaystyle =\int \frac{sec[(x)cos^2(x)-sin^2(x)]}{sin(x)+sec(x)}dx$

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

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

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

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

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

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

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

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

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

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

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

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

3. Hello, polymerase!

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

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

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

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

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

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

4. Originally Posted by Soroban
Hello, polymerase!

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

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

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

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

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

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

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

6. I just tripped over the most amazing approach . . .

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

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

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

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

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

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

Is that scary enough?

7. Originally Posted by Soroban

Is that scary enough?
Terrifying, Soroban, just terrifying!

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

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

10. Originally Posted by Krizalid
Well anyway to contemplate $\displaystyle u'$ it wasn't necessary to apply the product rule, just note $\displaystyle \sin x\cos x = \frac{{\sin 2x}} {2}$ 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...