If $\int\sec 2x\,dx= f[g(x)]+C$, then: |
$f(x)=\log|x|,g(x)=\tan(\frac{π}{4}-x)$ $f(x)=\log|x|,g(x)=\cot(\frac{π}{4}-x)$ $f(x)=\frac{1}{2}\log|x|,g(x)=\tan(\frac{π}{4}-x)$ $f(x)=\frac{1}{2}\log|x|,g(x)=\cot(\frac{π}{4}-x)$ |
$f(x)=\frac{1}{2}\log|x|,g(x)=\cot(\frac{π}{4}-x)$ |
$\int\sec 2x\,dx=\frac{1}{2}\frac{2\sec 2x(\sec 2x+\tan 2x)}{(\sec 2x+\tan 2x)}dx$ $=\frac{1}{2}\log|\sec 2x+\tan 2x|+C=\frac{1}{2}\log|\frac{1+\sin 2x}{\cos 2x}|$ $=\frac{1}{2}\log|\frac{1+\cos(\frac{π}{2}-2x)}{\sin(\frac{π}{2}-2x)}|+C=\frac{1}{2}\log|\cot(\frac{π}{4}-x)|+C$ $⇒f(x)=\frac{1}{2}\log|x|$ and $g(x)=\cot(\frac{π}{4}-x)$ |