Question

$\int se{c}^{3}xdx=$

• A. $\left(\frac{1}{2}\right)\left[secx\tan x-\log \left|secx+\tan x\right|\right]+c$
• B. $\left(\frac{1}{2}\right)\left[secx\tan x+\log \left|secx-\tan x\right|\right]+c$
• C. $\left(\frac{1}{2}\right)\left[secx\tan x-\log \left|secx-\tan x\right|\right]+c$
• D. $\left(\frac{1}{2}\right)\left[secx\tan x+\log \left|secx+\tan x\right|\right]+c$

Answer 1

The correct option is D

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

Explanation for correct option

Evaluate the integral.

Consider the given Equation as

$$\begin{gathered} I=\int se{c}^{3}xdx \\ I=\int secxse{c}^{2}xdx \end{gathered}$$

We know that

$$\begin{gathered} se{c}^{2}x=1+{\tan }^{2}x \\ secx=\sqrt{1+{\tan }^{2}x} \end{gathered}$$

Then the integral becomes,

$I=\int \left(\sqrt{1+{\tan }^{2}x}\right)se{c}^{2}xdx$

Let us assume that,

$$\begin{gathered} \tan x=t \\ se{c}^{2}xdx=dt \end{gathered}$$

Substitute the above value in integral

$I=\int \left(\sqrt{1+t^2}\right)dt$

We know that

$\int \left(\sqrt{a^2+x^2}\right)dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log \left|x+\sqrt{x^2+a^2}\right|+c$

Then,

$$\begin{gathered} I=\int \left(\sqrt{1+t^2}\right)dt \\ \Rightarrow I=\frac{1}{2}t\sqrt{1+t^2}+\frac{1}{2}\log \left|t+\sqrt{t^2+1}\right|+c \\ \Rightarrow I=\frac{1}{2}\left(\tan x\right)\sqrt{1+{\tan }^{2}x}+\frac{1}{2}\log \left|\tan x+\sqrt{{\tan }^{2}x+1}\right|+c \\ \Rightarrow I=\frac{1}{2}\tan x.secx+\frac{1}{2}\log \left|\tan x+secx\right|+c\left[\text{Where},\sqrt{1+{\tan }^{2}x}=secx\right] \end{gathered}$$

Hence, the correct answer is Option D.