$\displaystyle \int \frac{\sec\:x\;\cos\:2x}{\sin\;x+\sec\;x}\;dx$
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.
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: .$\displaystyle \ln|\sin2x + 2| + C$
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?