Question

$\int \left\{\frac{\left(secx\right)}{\left(secx+\tan x\right)}\right\}dx=$

• A. $\tan x-secx+c$
• B. $\log \left(1+secx\right)+c$
• C. $secx+\tan x+c$
• D. $\log \sin x+\log \cos x+c$

Answer 1

The correct option is A

$\tan x-secx+c$

Explanation for correct option:

Evaluate $\int \left\{\frac{\left(secx\right)}{\left(secx+\tan x\right)}\right\}dx$

consider the given equation as

$I=\int \left\{\frac{\left(secx\right)}{\left(secx+\tan x\right)}\right\}dx$

Multiply and divided by $secx-\tan x$ in the above Equation

$$\begin{gathered} I=\int \left\{\frac{\left(secx\right)\left(secx-\tan x\right)}{\left(secx+\tan x\right)\left(secx-\tan x\right)}\right\}dx \\ \Rightarrow I=\int \left\{\frac{\left(se{c}^{2}x-secx.\tan x\right)}{\left(se{c}^{2}x-{\tan }^{2}x\right)}\right\}dx \\ \Rightarrow I=\int \left(se{c}^{2}x-secx.\tan x\right)dx \\ \Rightarrow I=\tan x-secx+c \end{gathered}$$

Hence, the correct answer is Option A.