Mathematics

The value of the integral $\displaystyle \int\limits_0^1 {\dfrac{{{x^3}}}{{1 + {x^8}}}\,\,dx}$ is

$\frac{\pi }{{16}}$

SOLUTION
$\displaystyle \int _{ 0 }^{ 1 }{ \frac { { x }^{ 3 } }{ 1+{ x }^{ 8 } } dx }$

$=\displaystyle \int _{ 0 }^{ 1 }{ \frac { { x }^{ 3 } }{ 1+({ x }^{ 4 })^{ 2 } } dx }$

$=\displaystyle \dfrac { 1 }{ 4 } \int _{ 0 }^{ 1 }{ \frac { 1 }{ 1+(t)^{ 2 } } dt\quad (substituting\quad t={ x }^{ 4 }\Rightarrow dt=4x^3dx) }$

$=\dfrac { 1 }{ 4 } { \left[ { tan }^{ -1 }(t) \right] }_{ 0 }^{ 1 }$

$=\dfrac 14\left[\dfrac{\pi}{4}-0\right]$

$=\dfrac { \pi }{ 16 }$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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