Question

evaluate using substitution method ,integral (tan^-1x)^3/1+x^2 dx​

Answer 1

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

Formula Used :-

$\red{ \boxed{ \sf{ \dfrac{d}{dx} {tan}^{ - 1}x \: = \: \frac{1}{1 + {x}^{2} } }}}$

$\red{ \boxed{ \sf{\displaystyle\int\tt {x}^{n} \: dx \: = \: \frac{ {x}^{n + 1} }{n + 1} + c}}}$

Now,

Given Integral is

$\rm :\longmapsto\:\displaystyle\int\tt \dfrac{ {( {tan}^{ - 1} x)}^{3} }{1 + {x}^{2} } \: dx$

$\rm :\longmapsto\:Let \: I \: = \: \displaystyle\int\tt \dfrac{ {( {tan}^{ - 1} x)}^{3} }{1 + {x}^{2} } \: dx$

Now, we use method of Substitution.

$\red{\rm :\longmapsto\:Let \: {tan}^{ - 1}x \: = \: y}$

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

$\red{\rm :\longmapsto\:\dfrac{d}{dx} \: {tan}^{ - 1}x \: = \dfrac{d}{dx}\: y}$

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

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

So, on substituting all these values, in given integral, we get

$\rm :\longmapsto\:I \: = \: \displaystyle\int\tt {y}^{3} \: dy$

$\rm \: \: = \: \dfrac{ {y}^{3 + 1} }{3 + 1} + c$

$\rm \: \: = \: \dfrac{ {y}^{4} }{4} + c$

$\rm \: \: = \: \dfrac{ {( {tan}^{ - 1}x) } \: ^{4} }{4} + c$

Hence,

$\rm :\longmapsto\:\displaystyle\int\tt \dfrac{ {( {tan}^{ - 1} x)}^{3} }{1 + {x}^{2} } \: dx = \: \dfrac{ {( {tan}^{ - 1}x) } \: ^{4} }{4} + c$

Additional Information :-

Let take few more examples of Substitution

Example 1.

Evaluate

$\rm :\longmapsto\:\displaystyle\int\tt \dfrac{ {e}^{ {tan}^{ - 1} x} }{1 + {x}^{2} } \: dx$

Solution :-

$\red{\rm :\longmapsto\:Let \: {tan}^{ - 1}x \: = \: y}$

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

$\red{\rm :\longmapsto\:\dfrac{d}{dx} \: {tan}^{ - 1}x \: = \dfrac{d}{dx}\: y}$

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

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

So, On substituting all these values in above integral, we get

$\rm \: \: = \: \displaystyle\int\tt {e}^{y} \: dy$

$\rm \: \: = \: {e}^{y} \: + \: c$

$\red{\bigg \{ \because \: \displaystyle\int\tt {e}^{x}dx = {e}^{x} + c \bigg \}}$

$\rm \: \: = \: {e}^{ {tan}^{ - 1}x } \: + \: c$

Hence,

$\rm :\longmapsto\:\displaystyle\int\bf \dfrac{ {e}^{ {tan}^{ - 1} x} }{1 + {x}^{2} } \: dx = {e}^{ {tan}^{ - 1} x} + c$