Question

Evaluate $\int \frac{dx}{3\cos x+4\sin x+6}$

Answer 1

$\int \frac{dx}{3\cos x+4\sin x+6}$

As we know that,

$\cos x=\frac{1-{\tan }^2\left(\frac{x}{2}\right)}{1+{\tan }^2\left(\frac{x}{2}\right)}\text{and}\sin x=\frac{2\tan \left(\frac{x}{2}\right)}{1+{\tan }^2\left(\frac{x}{2}\right)}$

$\therefore \int \frac{dx}{3\cos x+4\sin x+6}=\int \frac{dx}{3\frac{1-{\tan }^2\left(\frac{x}{2}\right)}{1+{\tan }^2\left(\frac{x}{2}\right)}+4\frac{2\tan \left(\frac{x}{2}\right)}{1+{\tan }^2\left(\frac{x}{2}\right)}+6}$

$=\int \frac{1+{\tan }^2\left(\frac{x}{2}\right)dx}{3(1-{\tan }^2\left(\frac{x}{2}\right))+8\tan \left(\frac{x}{2}\right)+6(1+{\tan }^2\left(\frac{x}{2}\right))}$

$\Rightarrow =\int \frac{1+{\tan }^2\left(\frac{x}{2}\right)dx}{3-3{\tan }^2\left(\frac{x}{2}\right)+8\tan \left(\frac{x}{2}\right)+6+6{\tan }^2\left(\frac{x}{2}\right)}$

$\Rightarrow =\int \frac{{\sec }^2\left(\frac{x}{2}\right)dx}{3{\tan }^2\left(\frac{x}{2}\right)+8\tan \left(\frac{x}{2}\right)+9}[\because 1+{\tan }^2\theta ={\sec }^2\theta ]$

Let $\tan \left(\frac{x}{2}\right)=u$

$\Rightarrow \frac{1}{2}{\sec }^2\left(\frac{x}{2}\right)dx=du$

$\Rightarrow {\sec }^2\left(\frac{x}{2}\right)dx=2du$

$\therefore \int \frac{{\sec }^2\left(\frac{x}{2}\right)dx}{3{\tan }^2\left(\frac{x}{2}\right)+8\tan \left(\frac{x}{2}\right)+9}=\int \frac{2du}{3u^2+8u+9}$

$\Rightarrow =\int \frac{2du}{3(u^2+\frac{8}{3}u+3)}$

$\Rightarrow =\frac{2}{3}\int \frac{du}{{(u+\frac{4}{3})}^2-\frac{16}{9}+3}$

$\Rightarrow =\frac{2}{3}\int \frac{du}{{(u+\frac{4}{3})}^2+\frac{11}{9}}$

$\Rightarrow =\frac{2}{3}\int \frac{du}{{(u+\frac{4}{3})}^2+{\left(\frac{\sqrt{11}}{3}\right)}^2}$

Now the integral is in the form of $\int \frac{dx}{x^2+a^2}$.

$\therefore \frac{2}{3}\int \frac{du}{{(u+\frac{4}{3})}^2+{\left(\sqrt{\frac{11}{9}}\right)}^2}=\frac{2}{3}\times \frac{3}{\sqrt{11}}{\tan }^{-1}\left(\frac{u+\frac{4}{3}}{\sqrt{\frac{11}{9}}}\right)+c[\because \int \frac{dx}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}\frac{x}{a}+c]$

$\Rightarrow =\frac{2}{\sqrt{11}}{\tan }^{-1}\left(\frac{u+\frac{4}{3}}{\sqrt{\frac{11}{9}}}\right)+c$

As $u=\tan \frac{x}{2}$,

$\therefore \frac{2}{\sqrt{11}}{\tan }^{-1}\left(\frac{u+\frac{4}{3}}{\sqrt{\frac{11}{9}}}\right)+c=\frac{2}{\sqrt{11}}{\tan }^{-1}\left(\frac{\tan \frac{x}{2}+\frac{4}{3}}{\sqrt{\frac{11}{9}}}\right)+c$

Hence the required answer is $\int \frac{dx}{3\cos x+4\sin x+6}=\frac{2}{\sqrt{11}}{\tan }^{-1}\left(\frac{\tan \frac{x}{2}+\frac{4}{3}}{\sqrt{\frac{11}{9}}}\right)+c$.