Question

$\int \frac{\sin 2x}{4{\cos }^{2}x+9{\sin }^{2}x}dx=$

• A. $\frac{1}{13}\log (4{\cos }^{2}x+9{\sin }^{2}x)+C$
• B. $\frac{1}{12}\log (12{\tan }^{2}x+9)+C$
• C. $\frac{1}{5}\log (4{\cos }^{2}x+9{\sin }^{2}x)+C$
• D. $\log (4{\cos }^{2}x+9{\sin }^{2}x)+C$

Answer 1

The correct option is C $\frac{1}{5}\log (4{\cos }^{2}x+9{\sin }^{2}x)+C$

Let $I=\int \frac{\sin 2x}{4{\cos }^{2}x+9{\sin }^{2}x}dx$

$=\int \frac{2\sin x\cos x}{4+5{\sin }^{2}x}dx$

Put $4+5{\sin }^{2}x=t$

$\Rightarrow 10\sin x\cos xdx=dt$

$=\frac{1}{5}\int \frac{dt}{t}$

$=\frac{1}{5}\log |t|+C$

$I=\frac{1}{5}\log (4+5{\sin }^{2}x)+C$

$\Rightarrow I=\frac{1}{5}\log (4{\cos }^{2}x+9{\sin }^{2}x)+C$