Question

If $\int {\cos }^{2}3xco\sec 3xdx=\frac{1}{3}\left(\log |f\right(x)|+g(x\left)|\right)+C$, then

• A. $f\left(x\right)=\tan \frac{3x}{2}$
• B. $f\left(x\right)=\cot 3x$
• C. $g\left(x\right)=\sin 3x$
• D. $g\left(x\right)=\cos 3x$

Answer 1

Answer: (A), (D)

$f\left(x\right)=\tan \frac{3x}{2}$
$g\left(x\right)=\cos 3x$

Let $I=\int {\cos }^{2}3xco\sec 3xdx$

$=\int \left({1-\sin }^{2}3x\right)co\sec 3xdx$

$=\int (\frac{1}{\sin 3x}-\sin 3x)dx$

$=\int \frac{1}{\sin 3x}dx-\int \sin 3xdx$

$=-\frac{1}{3}\log |co\sec 3x+\cot 3x|+\cos 3x+C$

$=\frac{1}{3}\log \left|\frac{1}{co\sec 3x+\cot 3x}\right|+\cos 3x+C$

$\to f\left(x\right)=\frac{1}{co\sec 3x+\cot 3x}=\frac{\sin 3x}{1+\cos 3x}=\frac{2\sin \frac{3x}{2}\cos \frac{3x}{2}}{{\cos }^2\frac{3x}{2}}=\tan \frac{3x}{2}$

$\to g\left(x\right)=\cos 3x$