Mathematics - Indefinite Integration Question with Solution | TestHub

$\mathrm{I}=\int \frac{1}{1+3{\sin }^2\mathrm{x}+8{\cos }^2\mathrm{x}}\mathrm{dx}$ Dividing the numerator and denominator by ${\cos }^2\mathrm{x},$ we get $$I=\int \frac{{\sec }^{2}x}{{\sec }^{2}x+3{\tan }^{2}x+8}dx=\int \frac{{\sec }^{2}x}{4{\tan }^{2}x+9}dx$$ Putting $\tan x=t\Rightarrow {\sec }^{2}xdx=dt,$ we get $\mathrm{I}=\int \frac{\mathrm{dt}}{4{\mathrm{t}}^2+9}=\frac{1}{4}\int \frac{\mathrm{dt}}{{\mathrm{t}}^2+(3/2{)}^2}=\frac{1}{4}\times \frac{1}{3/2}{\tan }^{-1}\left(\frac{\mathrm{t}}{3/2}\right)+\mathrm{C}$ $\Rightarrow \mathrm{I}=\frac{1}{6}{\tan }^{-1}\left(\frac{2\mathrm{t}}{3}\right)+\mathrm{C}=\frac{1}{6}{\tan }^{-1}\left(\frac{2\tan \mathrm{x}}{3}\right)+\mathrm{C}$