Question

integral of x³sin(tan-¹x⁴)/(1+x⁸)​

Answer 1

Answer:

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Answer 2

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$1. \: \: \: \boxed{ \sf \displaystyle\int \tt \:sinx = - cosx + c}$

$2. \: \: \: \boxed{ \rm \: \dfrac{d}{dx} {tan}^{ - 1} x = \dfrac{1}{1 + {x}^{2} } }$

$3. \: \: \: \boxed{ \rm \: \dfrac{d}{dx} {x}^{n} = n {x}^{n - 1} }$

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

$\sf \longmapsto \displaystyle\int \rm \: \dfrac{ {x}^{3} {sin\bigg( {tan}^{ - 1} {x}^{4} \bigg) } }{1 + {x}^{8} } dx$

As we know that,

• By using substitution method, we get.

$\rm :\longmapsto\: {tan}^{ - 1} {x}^{4} = y$

$\rm :\longmapsto\: \rm \: Differentiate \: w.r.t \: x, \: we \: get$

$\rm :\longmapsto\:\dfrac{1}{1 + {x}^{8} } \times {4x}^{3} = \dfrac{dy}{dx}$

$\rm :\implies\:\dfrac{ {x}^{3} }{ 1 + {x}^{8} } dx = \dfrac{dy}{4}$

On substituting these values in given integral, we get

$\sf \longmapsto \displaystyle \: = \: \dfrac{1}{4} \int \tt \: siny \: dy$

$\rm :\longmapsto\: \rm \: = - \: \dfrac{1}{4} \: cosy \: + \: c$

(On substituting back the value of y, we get)

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

Additional Information :-

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

$2. \: \: \: \boxed{ \sf \displaystyle\int \tt \:k \: dx = k x\: + \: c}$

$3. \: \: \: \boxed{ \sf \displaystyle\int \tt \:cosx \: dx \: = \: sinx \: + \: c}$

$4. \: \: \: \boxed{ \sf \displaystyle\int \tt \:\dfrac{1}{x} dx = log(x) + c }$

$5. \: \: \: \boxed{ \sf \displaystyle\int \tt \: {e}^{x}dx = {e}^{x} + c}$

$6. \: \: \: \boxed{ \sf \displaystyle\int \tt \: {sec}^{2}x = tanx + c }$

$7. \: \: \: \boxed{ \sf \displaystyle\int \tt \: {cosec}^{2} x = - cotx + c}$

$8. \: \: \: \boxed{ \sf \displaystyle\int \tt \:secx \: tanx \: dx = secx + c}$

$9. \: \: \: \boxed{ \sf \displaystyle\int \tt \:cosecx \: cotx \: dx = - cosecx + c}$