Question

$\int \frac{1}{3\sin x+4\cos x}dx$ equal, where $\tan \alpha =\frac{4}{3}$

• A. $\frac{1}{5}ln\left|\tan (\frac{x}{2}+\frac{\alpha }{2})\right|+K$
• B. $\frac{1}{5}ln\left|\tan (x+\alpha )\right|+K$
• C. $\frac{1}{5}ln\left|\tan (\frac{x}{2}+\frac{\pi }{2}-\frac{\alpha }{2})\right|+K$
• D. $\frac{1}{5}ln\left|\tan (\frac{x}{2}-\frac{\alpha }{2})\right|+K$

Answer 1

Answer: (A)

$\frac{1}{5}ln\left|\tan (\frac{x}{2}+\frac{\alpha }{2})\right|+K$

$\int \frac{1}{3\sin x+4\cos x}dx=\int 5\frac{1}{(\frac{3}{5}\sin x+\frac{4}{5}\cos x)}dx$
$=\frac{1}{5}\int \frac{1}{\sin x\cdot \cos \alpha +\cos x\cdot \sin \alpha }dx$ $\tan \alpha =\frac{4}{3}$
$=\int \frac{1}{\sin (x+\alpha )}dx$ $\sin \alpha =\frac{4}{5}$
$=\frac{1}{5}\int cose(x+\alpha )dx$ $\cos \alpha =\frac{3}{5}$
$=\frac{1}{5}ln\left|\tan (\frac{\pi }{2}+\frac{d}{2})\right|+k$