Question

Answer 1

$\mathrm{Let}I=\int \frac{1}{4{\sin }^{2}x+4\sin x·\cos x+5{\cos }^{2}x}dx$

Dividing numerator and denominator by cos²x we get
$$\begin{gathered} I=\int \frac{{\sec }^{2}x}{4{\tan }^{2}x+4\tan x+5}dx \\ \mathrm{Putting}\tan x=t \\ \Rightarrow {\sec }^{2}xdx=dt \\ \therefore I=\int \frac{dt}{4t^2+4t+5} \\ =\frac{1}{4}\int \frac{dt}{t^2+t+{\displaystyle \frac{5}{4}}} \\ =\frac{1}{4}\int \frac{dt}{t^2+t+{\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{4}}+{\displaystyle \frac{5}{4}}} \\ =\frac{1}{4}\int \frac{dt}{{\left(t+{\displaystyle \frac{1}{2}}\right)}^2+1^2} \\ =\frac{1}{4}\times {\tan }^{-1}\left(t+\frac{1}{2}\right)+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left[\because \int \frac{1}{x^{\mathit{2}}\mathit{+}{a}^{\mathit{2}}}dx\mathit{=}\frac{1}{a}{\tan }^{-1}\frac{x}{a}+C\right] \\ =\frac{1}{4}{\tan }^{-1}\left(\frac{2t+1}{2}\right)+C \\ =\frac{1}{4}{\tan }^{-1}\left(\frac{2\tan x+1}{2}\right)+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left[\because \mathit{}t=\tan \mathit{}x\right] \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}\frac{1}{4}{\tan }^{-1}\mathit{}\left(\tan x+\frac{1}{2}\right)+C\mathit{}\mathit{}\mathit{}\mathit{} \end{gathered}$$