Question

If $\int \frac{\sin 2x}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}dx=k\cdot \ln |a^2{\cos }^{2}x+b^2{\sin }^{2}x|+C,$ where $a\ne b$, then $k$ is equal to
(where $C$ is integration constant)

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

Answer 1

The correct option is A $\frac{1}{b^2-a^2}$

$I=\int \frac{\sin 2x}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}dx$
Put $a^2{\cos }^{2}x+b^2{\sin }^{2}x=t$
$\Rightarrow (-a^2\sin 2x+b^2\sin 2x)dx=dt\Rightarrow \sin 2xdx=\frac{dt}{b^2-a^2}\Rightarrow I=\frac{1}{b^2-a^2}\int \frac{dt}{t}\Rightarrow I=\frac{1}{b^2-a^2}\cdot \ln |t|+C\Rightarrow I=\frac{1}{b^2-a^2}\cdot \ln |a^2{\cos }^{2}x+b^2{\sin }^{2}x|+C\therefore k=\frac{1}{b^2-a^2}$