Question

\[\int\frac{1}{4 \sin^2 x + 4 \sin x \cos x + 5 \cos^2 x} \text{ dx }\]

Answer 1

\[\text{ Let  I } = \int\frac{1}{4 \sin^2 x + 4 \sin x \cdot \cos x + 5 \cos^2 x}dx\]

Dividing numerator and denominator by cos²x we get

\[I = \int\frac{\sec^2 x}{4 \tan^2 x + 4 \tan x + 5}dx\]

\[\text{ Putting tan x = t}\]

\[ \Rightarrow \text{ sec}^2 \text{ x  dx = dt }\]

\[ \therefore I = \int\frac{dt}{4 t^2 + 4t + 5}\]

\[ = \frac{1}{4}\int\frac{dt}{t^2 + t + \frac{5}{4}}\]

\[ = \frac{1}{4}\int\frac{dt}{t^2 + t + \frac{1}{4} - \frac{1}{4} + \frac{5}{4}}\]

\[ = \frac{1}{4}\int\frac{dt}{\left( t + \frac{1}{2} \right)^2 + 1^2}\]

\[ = \frac{1}{4} \times \tan^{- 1} \left( t + \frac{1}{2} \right) + C.......... \left[ \because \int\frac{1}{x^2 + a^2}dx = \frac{1}{a} \tan^{- 1} \frac{x}{a} + C \right]\]

\[ = \frac{1}{4} \tan^{- 1} \left( \frac{2t + 1}{2} \right) + C\]

\[ = \frac{1}{4} \tan^{- 1} \left( \frac{2 \tan x + 1}{2} \right) + C...........\left[ \because t = \tan x \right]\]

\[ = \frac{1}{4} \tan^{- 1} \left( \tan x + \frac{1}{2} \right) + C\]