Question

$\int \left\{\frac{\left(secx\cos ecx\right)}{\left(2cotx-secx\cos ecx\right)}\right\}dx=$

• A. $\log \left|secx+\tan x\right|+c$
• B. $\log \left|secx+\cos ecx\right|+c$
• C. $\left(\frac{1}{2}\right)\log \left|sec2x+\tan 2x\right|+c$
• D. $\log \left|sec2x+\cos ec2x\right|+c$
• E. $\log \left|sec2x.\cos ec2x\right|+c$

Answer 1

The correct option is C

$\left(\frac{1}{2}\right)\log \left|sec2x+\tan 2x\right|+c$

Explanation for correct option:

Evaluate : $\int \left\{\frac{\left(secx\cos ecx\right)}{\left(2cotx-secx\cos ecx\right)}\right\}dx$

Consider the given Equation as

$$\begin{gathered} I=\int \left\{\frac{\left(secx\cos ecx\right)}{\left(2cotx-secx\cos ecx\right)}\right\}dx \\ I=\int \frac{{\displaystyle \frac{1}{\sin x.\cos x}}dx}{{\displaystyle \frac{2\cos x}{\sin x}}-{\displaystyle \frac{1}{\cos x.\sin x}}} \\ I=\int \frac{1}{2{\cos }^2-1}dx \\ I=\int \frac{1}{\cos 2x}dx\left[\text{where},2{\cos }^2-1=\cos 2x\right] \\ I=\int sec2xdx\left[\text{where},\frac{1}{\cos 2x}=sec2x\right] \\ I=\left(\frac{1}{2}\right)\log \left|sec2x+\tan 2x\right|+c \end{gathered}$$

Hence the correct answer is Option C.