Question

Evaluate

$\int \frac{14secx+8cosx}{2secx+3cosx}dx$

Answer 1

$\begin{array}{c}{\int }^{}\frac{14\sec \left(x\right)+8\cos \left(x\right)}{2\sec \left(x\right)+3\cos \left(x\right)}dx=2{\int }^{}\frac{7\sec \left(x\right)+4\cos \left(x\right)}{2\sec \left(x\right)+3\cos \left(x\right)}dx\\ =2{\int }^{}{\sec }^2\left(x\right)\cdot \frac{7{\tan }^2\left(x\right)+11}{\left({\tan }^2\left(x\right)+1\right)\left(2{\tan }^2\left(x\right)+5\right)}dx\\ substituteu=\tan x,\\ ={\int }^{}\frac{7u^2+11}{\left(u^2+1\right)\left(2u^2+5\right)}dx\\ =\frac{13ta{n}^{-1}\left(\frac{\sqrt{2}u}{\sqrt{5}}\right)}{3\cdot \sqrt{2}\cdot \sqrt{5}}+\frac{4ta{n}^{-1}\left(u\right)}{3}\\ =2{\int }^{}\frac{7\sec \left(x\right)+4\cos \left(x\right)}{2\sec \left(x\right)+3\cos \left(x\right)}dx=\frac{13\cdot \sqrt{2}ta{n}^{-1}\left(\frac{\sqrt{2}\tan \left(x\right)}{\sqrt{5}}\right)}{3\cdot \sqrt{5}}+\frac{8ta{n}^{-1}\left(\tan \left(x\right)\right)}{3}\\ =\frac{13\cdot \sqrt{2}ta{n}^{-1}\left(\frac{\sqrt{2}\tan \left(x\right)}{\sqrt{5}}\right)}{3\cdot \sqrt{5}}+\frac{8x}{3}+C\end{array}$