Question

` ∫  {sin 2x} /{a cos^2  x  + b sin^2  x }  ` dx 

Answer 1

\[\text{Let I} = \int\frac{\sin 2x}{a \cos^2 x + b \sin^2 x}dx\]
\[ = \int\frac{\sin 2x}{a\left( 1 - \sin^2 x \right) + b \sin^2 x} dx\]
\[ = \int\frac{\sin 2x}{\left( b - a \right) \sin^2 x + a} dx\]

`  "Putting "     s   "in" ^2 x = t `
\[ \Rightarrow 2\sin x . \cos x = \frac{dt}{dx}\]
\[ \Rightarrow \sin 2x = \frac{dt}{dx}\]
\[ \Rightarrow \text{sin 2x dx }= dt\]
\[ \therefore I = \int\frac{1}{\left( b - a \right)t + a}dt\]
\[ = \frac{1}{\left( b - a \right)} \text{ln }\left| \left( b - a \right)t + a \right| + C \left[ \because \int\frac{1}{ax + b}dx = \frac{1}{a}\text{ln}\left| ax + b \right| + C \right]\]
\[ = \frac{1}{\left( b - a \right)} \text{ln }\left| \left( b - a \right) \sin^2 x + a \right| + C \left[ \because t = \sin^2 x \right]\]
\[ = \frac{1}{\left( b - a \right)} \text{ln }\left| b \sin^2 x + a\left( 1 - \sin^2 x \right) \right| + C\]
\[ = \frac{1}{\left( b - a \right)} \text{ln} \left| b \sin^2 x + a \cos^2 x \right| + C\]