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

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

$$\text{ Putting tan x = t}$$

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

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

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

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

$$=\frac{1}{4}\int \frac{dt}{{(t+\frac{1}{2})}^2+1^2}$$

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

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

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

$$=\frac{1}{4}{\tan }^{-1}(\tan x+\frac{1}{2})+C$$