Question

solve the question step by step​

Answer 1

Solution:

Given Integral:

$\displaystyle \rm \longrightarrow I = \int \sqrt{1 + \sec(x) } \: dx$

Can be written as:

$\displaystyle \rm \longrightarrow I = \int \sqrt{ \dfrac{ \{1 + \sec(x) \} \{1 - \sec(x) \}}{1 - \sec(x) } } \: dx$

$\displaystyle \rm \longrightarrow I = \int \sqrt{ \dfrac{ \{\sec(x) + 1\} \{\sec(x) - 1\}}{\sec(x) - 1}} \: dx$

$\displaystyle \rm \longrightarrow I = \int \sqrt{ \dfrac{\sec^{2} (x) - 1}{\sec(x) - 1}} \: dx$

$\displaystyle \rm \longrightarrow I = \int \sqrt{ \dfrac{\tan^{2} (x)}{\sec(x) - 1}} \: dx$

$\displaystyle \rm \longrightarrow I = \int \dfrac{\tan(x)}{ \sqrt{\sec(x) - 1}} \: dx$

Let us assume that:

$\rm \longrightarrow u = \sqrt{ \sec(x) - 1}$

$\rm \longrightarrow {u}^{2} = \sec(x) - 1$

$\rm \longrightarrow 2u \dfrac{du}{dx} = \sec(x) \tan(x)$

$\rm \longrightarrow dx =\dfrac{2u \:du}{ \sec(x) \tan(x)}$

So, the integral changes to:

$\displaystyle \rm \longrightarrow I = \int \dfrac{\tan(x)}{u} \times \dfrac{2u \: du}{ \sec(x) \tan(x) }$

$\displaystyle \rm \longrightarrow I = 2\int\dfrac{ du}{ \sec(x)}$

$\displaystyle \rm \longrightarrow I = 2\int\dfrac{ du}{ {u}^{2} + {1}^{2} }$

$\displaystyle \rm \longrightarrow I = 2 \tan^{ - 1}(u) + C$

Substituting back the value of u, we get:

$\displaystyle \rm \longrightarrow I = 2 \tan^{ - 1}( \sqrt{ \sec(x) - 1 } ) + C$

Therefore:

$\displaystyle \rm \longrightarrow \int \sqrt{1 + \sec(x) } \: dx = 2 \tan^{ - 1}( \sqrt{ \sec(x) - 1 } ) + C$

★ Which is our required answer.

Learn More:

$\boxed{\begin{array}{c|c}\bf f(x)&\bf\displaystyle\int\rm f(x)\:dx\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \rm k&\rm kx+C\\ \\ \rm sin(x)&\rm-cos(x)+C\\ \\ \rm cos(x)&\rm sin(x)+C\\ \\ \rm{sec}^{2}(x)&\rm tan(x)+C\\ \\ \rm{cosec}^{2}(x)&\rm-cot(x)+C\\ \\ \rm sec(x)\ tan(x)&\rm sec(x)+C\\ \\ \rm cosec(x)\ cot(x)&\rm-cosec(x)+C\\ \\ \rm tan(x)&\rm log(sec(x))+C\\ \\ \rm\dfrac{1}{x}&\rm log(x)+C\\ \\ \rm{e}^{x}&\rm{e}^{x}+C\\ \\ \rm x^{n},n\neq-1&\rm\dfrac{x^{n+1}}{n+1}+C\end{array}}$