নির্দিষ্ট যোগজ

The value of the integral 01x31+x8dx\displaystyle \int\limits_0^1 {\dfrac{{{x^3}}}{{1 + {x^8}}}\,\,dx} is

হানি নাটস

01x31+x8dx\displaystyle \int _{ 0 }^{ 1 }{ \frac { { x }^{ 3 } }{ 1+{ x }^{ 8 } } dx }

=01x31+(x4)2dx=\displaystyle \int _{ 0 }^{ 1 }{ \frac { { x }^{ 3 } }{ 1+({ x }^{ 4 })^{ 2 } } dx }

=140111+(t)2dt(substitutingt=x4dt=4x3dx)=\displaystyle \dfrac { 1 }{ 4 } \int _{ 0 }^{ 1 }{ \frac { 1 }{ 1+(t)^{ 2 } } dt\quad (substituting\quad t={ x }^{ 4 }\Rightarrow dt=4x^3dx) }

=14[tan1(t)]01=\dfrac { 1 }{ 4 } { \left[ { tan }^{ -1 }(t) \right] }_{ 0 }^{ 1 }

=14[π40]=\dfrac 14\left[\dfrac{\pi}{4}-0\right]

=π16=\dfrac { \pi }{ 16 }

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