Question

integral of sec^x/✓(tan^2x+4) ​

Answer 1

Appropriate Question

$\rm :\longmapsto\:\displaystyle\int\tt \dfrac{ {sec}^{2} x}{ \sqrt{ {tan}^{2}x + 4 } } \: dx$

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm :\longmapsto\:\displaystyle\int\tt \dfrac{ {sec}^{2} x}{ \sqrt{ {tan}^{2}x + 4 } } \: dx$

To evaluate this integral, we use method of Substitution,

$\red{\rm :\longmapsto\:Put \: tanx = y}$

On differentiating both sides w. r. t. x, we get

$\red{\rm :\longmapsto \: \dfrac{d}{dx} \: tanx =\dfrac{d}{dx} y}$

$\red{\rm :\longmapsto \: {sec}^{2} x =\dfrac{dy}{dx} }$

$\red{\rm :\longmapsto \: {sec}^{2} xdx = dy}$

So, above given integral can be rewritten as

$\rm \: = \: \:\displaystyle\int\tt \frac{dy}{ \sqrt{ {y}^{2} + 4} }$

$\rm \: = \: \:\displaystyle\int\tt \frac{dy}{ \sqrt{ {y}^{2} + {2}^{2} } }$

We know that

$\boxed{ \bf{ \: \displaystyle\int\tt \frac{dx}{ \sqrt{ {x}^{2} + {a}^{2} } } = log \: | \: x \: + \: \sqrt{ {x}^{2} + {a}^{2} } \: | \: + \: c}}$

So, using this result, we get

$\rm \: = \: \:log \: | \: y \: + \: \sqrt{ {y}^{2} + {2}^{2} } \: | \: + \: c$

$\rm \: = \: \:log \: | \: y \: + \: \sqrt{ {y}^{2} + 4 } \: | \: + \: c$

On substituting the value of y = tanx, we get

$\rm \: = \: \:log \: | \: tanx \: + \: \sqrt{ {tan}^{2}x + 4 } \: | \: + \: c$

Hence,

$\rm \: \displaystyle\int\tt \frac{ {sec}^{2} x}{ \sqrt{ {tan}^{2} x + 4} } \: dx = \: \:log \: | \: tanx \: + \: \sqrt{ {tan}^{2}x + 4 } \: | \: + \: c$

Additional Information :-

$\boxed{ \bf{ \: \displaystyle\int\tt \frac{dx}{ \sqrt{ {x}^{2} - {a}^{2} } } = log \: | \: x \: + \: \sqrt{ {x}^{2} - {a}^{2} } \: | \: + \: c}}$

$\boxed{ \bf{ \: \displaystyle\int\tt \frac{dx}{ \sqrt{ {a}^{2} - {x}^{2} } } = {sin}^{ - 1} \frac{x}{a} \: + \: c}}$

$\boxed{ \bf{ \: \displaystyle\int\tt \frac{dx}{ {x}^{2} + {a}^{2}} = \frac{1}{a} \: {tan}^{ - 1} \frac{x}{a} \: + \: c}}$

$\boxed{ \bf{ \: \displaystyle\int\tt \frac{dx}{ {x}^{2} - {a}^{2}} = \frac{1}{2a} \: {log} \: \: \frac{x - a}{x + a} \: + \: c}}$