Question

How do u integrate dx/$sec^2$($tan^{-1}$x)​

Answer 1

$\red{\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese  \:  \: TO \: \: CALCULATE \:  \:  \maltese }}}}}$

$\sf \int \frac{dx}{ {sec}^{2}(tan {}^{ - 1} x)}$

$\green{\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese  \:  \:  KEY \: \: POINT \:  \: \maltese }}}}}$

$\large\mathfrak{Express \: \: sec^2 \text{}(tan^{-1} \text{x })\: \: in \: \: } \\ \large\mathfrak{terms \: \: of \: \: \text{x }\: \: onl \text{y} \: \: \text{And } } \\ \large \mathfrak{ solve \: \: after \: \: Expressing}$

$\color{magenta}\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese  \:  \: REQUIRED \: \: INFO \:  \:   \maltese }}}} \\ \\ \bigstar \: \bf {sec}^{2} \theta = 1 + {tan}^{2} \theta \\ \\ \bf \bigstar \: \int \frac{dx}{1 + {x}^{2} } = {tan}^{ - 1} x + C$

$\purple{\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese  \:  \:  SOLUTION \:  \:  \maltese }}}}}$

$\blue{\boxed{ \boxed{ \bigstar \bf Expressing \: in \: terms \: of \: x }}}$

$\huge\mathfrak{ \text{L}et} \\ \\ \bf {tan}^{ - 1} x = \alpha \implies \sf\tan\alpha = x \\ \\ SO\\ \\ \sf{sec}^{2} ( {tan}^{ - 1} x) = {sec}^{2} \alpha \\ \\ = \sf 1 + {tan}^{2} \alpha \\ \\ = \bf 1 + {x}^{2}\\ \implies \green{ \boxed{ \boxed{\bf{sec}^{2} ( {tan}^{ - 1} x) =1 + {x}^{2} }}}\:\:\: ...(i)$

$\huge\mathfrak{ \text{N}o \sf{w}} \\ \\ \sf \int \frac{dx}{ {sec}^{2} ( {} tan {}^{ - 1}x) } \\ \\= \sf \int \frac{dx}{1 + {x}^{2} }\:\:\:\:\bf[using\: Eq^n (i)] \\ \\ = \bf{tan}^{ - 1} x + C$

Which is the required Answer

Answer 2

Question:

$\huge \int \frac{dx}{ {sec}^{2}( {tan}^{ - 1} x) }$

Useful Formulas:

$\bf1) \: \: \sec {}^{2} \alpha = 1 + { \tan}^{2} \alpha \\ \\ \bf2) \: \: \int \frac{dx}{1 + {x}^{2} } = tan {}^{ - 1} x + c$

${\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \bigstar  \:  SOLUTION }}}}}$

Firstly,

Let

${tan}^{ - 1} x = \theta \implies \tan\theta = x \\ \\ \Large Now,\\ \\ {sec}^{2} ( {tan}^{ - 1} x) = {sec}^{2} \theta\\ \\ = 1 + {tan}^{2} \theta \\ \\ = 1 + {x}^{2}$

$\huge\mathfrak{ Finall\sf{y}} \\ \\ \int \frac{dx}{ {sec}^{2} ( {} tan {}^{ - 1}x) }\\\\ = \int \frac{dx}{1 + {x}^{2} } \\ \\ = \bf{tan}^{ - 1} x + C$