Question

Evaluate: $\int \frac{\sin 2x}{a^2+b^2{\sin }^{2}x}dx$

• A. $-\frac{1}{b^2}\log (a^2+b^2{\sin }^{2}x)+c$
• B. $-b^2\log (a^2+b^2{\sin }^{2}x)+c$
• C. $\frac{1}{b^2}\log (a^2+b^2{\sin }^{2}x)+c$
• D. $b^2\log (a^2+b^2{\sin }^{2}x)+c$

Answer 1

The correct option is C $\frac{1}{b^2}\log (a^2+b^2{\sin }^{2}x)+c$

$\int \frac{\sin 2x}{a^2+b^2{\sin }^{2}x}dx=\int \frac{2\sin x\cos x}{a^2+b^2{\sin }^{2}x}dx=\frac{1}{b^2}\int \frac{2b^2\sin x\cos x}{a^2+b^2{\sin }^{2}x}dx$

Let $t=a^2+b^2{\sin }^{2}xdt=0+b^{2}2\sin x\cos xdx$

Hence, Given Equation can be written as $\frac{1}{b^2}\int \frac{dt}{t}=\frac{1}{b^2}\log t+c=\frac{1}{b^2}\log (a^2+b^2{\sin }^{2}x)+c$