Question

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

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

Answer 1

The correct option is A $\frac{1}{b^2}\log |a^2+b^2{\sin }^{2}x|+c$

Let $I=\int \frac{\sin 2x}{a^2+b^2{\sin }^{2}x}dx$

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

But $\cos x\cdot dx=d\sin x$

$\therefore I=\int \frac{2\sin xd\sin x}{a^2+b^2{\sin }^{2}x}$

$=\int \frac{d{\sin }^{2}x}{a^2+b^2{\sin }^{2}x}$

Let ${\sin }^{2}x=t$

Then,

$I=\int \frac{dt}{a^2+b^{2}t}$

$=\frac{1}{b^2}\log (a^2+b^{2}t)+c$

$=\frac{1}{b^2}\log (a^2+b^2{\sin }^{2}x)+c$

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