Question

solve this integration quickly

Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \frac{ {x}^{n - 1} }{ \sqrt{ {a}^{n} + {x}^{n} } } \: dx$

To solve this integral, we use method of Substitution.

$\red{\rm :\longmapsto\:Put \: \sqrt{ {a}^{n} + {x}^{n} } = y}$

$\red{\rm :\longmapsto\: \: { {a}^{n} + {x}^{n} } = {y}^{2} }$

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

$\red{\rm :\longmapsto\: \: \dfrac{d}{dx}({ {a}^{n} + {x}^{n} }) = \dfrac{d}{dx} {y}^{2} }$

$\red{\rm :\longmapsto\: \: \dfrac{d}{dx}{ {a}^{n} + \dfrac{d}{dx}{x}^{n} } =2y \dfrac{dy}{dx}}$

$\red{\rm :\longmapsto\: \: {nx}^{n - 1} =2y \dfrac{dy}{dx}}$

$\red{\rm :\longmapsto\: {x}^{n - 1}dx = \dfrac{2y}{n}dy}$

Now, on substituting all the above values, we get

$\rm \: = \: \: \dfrac{2}{n} \displaystyle\int\rm \frac{y \: dy}{y}$

$\rm \: = \: \: \dfrac{2}{n} \displaystyle\int\rm dy$

$\rm \: = \: \: \dfrac{2}{n} \: y \: + \: c$

$\rm \: = \: \: \dfrac{2 \sqrt{ {a}^{n} + {x}^{n} } }{n} \ \: + \: c$

Hence,

$\rm :\longmapsto\:\displaystyle\int\bf \frac{ {x}^{n - 1} }{ \sqrt{ {a}^{n} + {x}^{n} } } \: dx$

is

$\bf \: = \: \: \dfrac{2 \sqrt{ {a}^{n} + {x}^{n} } }{n} \ \: + \: c$

Additional Information :-

$\displaystyle\int\rm sinx \: dx \: = \: - \: cosx \: + \: c$

$\displaystyle\int\rm cosx \: dx \: = \: \: sinx \: + \: c$

$\displaystyle\int\rm tanx \: dx \: = \: \: log \: secx \: + \: c$

$\displaystyle\int\rm cotx \: dx \: = \: \: log \: sinx \: + \: c$

$\displaystyle\int\rm cosecx \: dx \: = \: - \: log \: (cosecx - cotx)\: + \: c$

$\displaystyle\int\rm secx \: dx \: = \: \: log \: (secx + tanx)\: + \: c$