Question

Write a value of\[\int\frac{\sin 2x}{a^2 \sin^2 x + b^2 \cos^2 x} \text{ dx }\]

Answer 1

\[\text{ Let I} = \int \frac{\text{ sin   2x  dx}}{a^2 \sin^2 + b^2 \cos^2 x}\]
\[\text{ Let a}^2 \sin^2 x + b^2 \cos^2 x = t\]
\[ \Rightarrow \left[ a^2 \left( 2 \sin x \cos x \right) + b^2 \left( 2 \cos x \times - \sin x \right) \right]dx = dt\]
\[ \Rightarrow \left( a^2 - b^2 \right) \text{ sin 2x . dx} = dt\]
\[ \Rightarrow \text{ sin 2x dx }= \frac{dt}{a^2 - b^2}\]
\[ \therefore I = \frac{1}{a^2 - b^2}\int\frac{dt}{t}\]
\[ = \frac{1}{a^2 - b^2}\log t + C\]
\[ = \frac{1}{a^2 - b^2}\log \left(\text{  a} ^2 \sin^2 x + b^2 \cos^2 x \right) + C \left( \because t = a^2 \sin^2 x + b^2 \cos^2 x \right)\]