Question

Evaluate $\int \frac{\sec x}{2\cos x+5\sin x}dx$

• A. $\frac{1}{2}\log |(2\cos x+5\sin x)|+c$
• B. $\frac{1}{5}\log |(2+5\tan x)|+c$
• C. $\frac{1}{2}\log |(2+3\tan x)|+c$
• D. $\frac{1}{3}\log |(2+5\tan x)|+c$

Answer 1

The correct option is B $\frac{1}{5}\log |(2+5\tan x)|+c$

$\int \frac{secx}{2cosx+5sinx}dx=\int \frac{secx}{cosx(2+5tanx)}dx\cdot$

$\int \frac{se{c}^{2}x}{2+5tanx}dx\cdot$

$tanx=t$

$se{c}^{2}x\cdot dx=dt$

$\int \frac{dt}{2+5t}=\frac{1}{5}log(2+5t)+c$

By putting the value of ${}^{\prime }{t}^{\prime }$ we get,

$=\frac{1}{5}log\left|\right(2+5tanx\left)\right|+c\cdot$

$\int \frac{secx}{2cosx+5sinx}dx=\frac{1}{5}log\left|\right(2+5tanx\left)\right|+c\cdot$