Question

Integrate $\int \frac{\sec \left(\mathrm{x}\right)\tan \left(\mathrm{x}\right)}{3\sec \left(\mathrm{x}\right)+5}dx$.

Answer 1

Solve the given integral

Let, $I=\int \frac{\sec \left(\mathrm{x}\right)\tan \left(\mathrm{x}\right)}{3\sec \left(\mathrm{x}\right)+5}dx$

Multiply and divide by $3$ on RHS,

$\Rightarrow \mathrm{I}=\frac{1}{3}\int \frac{3\sec \left(\mathrm{x}\right)\tan \left(\mathrm{x}\right)}{3\sec \left(\mathrm{x}\right)+5}\mathrm{dx}$

Let, $3\sec \left(\mathrm{x}\right)+5=\mathrm{t};\left[\because \frac{d\sec \left(\mathrm{x}\right)}{d\mathrm{x}}=\sec \left(\mathrm{x}\right)\tan \left(\mathrm{x}\right),\frac{d\left(\mathrm{constant}\right)}{d\mathrm{x}}=0\right]$

Differentiate both sides,

$\Rightarrow 3\sec \left(\mathrm{x}\right)\tan \left(\mathrm{x}\right)\mathrm{dx}=\mathrm{dt}$

Replacing value of $dx$ in $I$,

$$\begin{gathered} \Rightarrow \mathrm{I}=\frac{1}{3}\int \frac{\mathrm{dt}}{\mathrm{t}}\left[\because \int \frac{\mathrm{dt}}{\mathrm{t}}=\ln \left(\mathrm{t}\right)+\mathrm{c}\right] \\ \Rightarrow \mathrm{I}=\frac{\ln \left(\mathrm{t}\right)}{3}+\mathrm{c} \\ \Rightarrow \mathrm{I}=\frac{\ln \left(3\sec \left(\mathrm{x}\right)+5\right)}{3}+\mathrm{C} \end{gathered}$$

Hence, $\int \frac{\sec \left(\mathrm{x}\right)\tan \left(\mathrm{x}\right)}{3\sec \left(\mathrm{x}\right)+5}dx$ is integrated as $\frac{\ln \left(3\sec \left(\mathrm{x}\right)+5\right)}{3}+C$.