If $\int\frac{dx}{5+4\sin x}=A\tan^{-1}(B\tan\frac{x}{2}+\frac{x}{3})+C$, then:
Answer & explanation
Correct answer: option 1
$\int\frac{dx}{5+4\sin x}=\int\frac{dx}{5+4(\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}})}$
$=\int\frac{\sec^2\frac{x}{2}}{5\tan^2\frac{x}{2}+8\tan\frac{x}{2}+5}=2\int\frac{dz}{5z^2+8z+5}$ $[Putting\,\tan\frac{x}{2}=z⇒\sec^2\frac{x}{2}dx=2\,dz]$
$=\frac{2}{5}\int\frac{dz}{z^2+1+\frac{8}{5}z}=\frac{2}{5}\int\frac{dx}{(z+\frac{4}{5})+(\frac{3}{5})^2}$
$=\frac{2}{3}.\frac{5}{3}\tan^{-1}(\frac{z+\frac{4}{5}}{\frac{3}{5}})+C=\frac{2}{3}\tan^{-1}(\frac{5\tan\frac{x}{2}+4}{3})+C$
$∴A=\frac{2}{3}$ and $B=\frac{5}{3}$