one more q (1 Viewer)

5uckerberg · Jun 11, 2022

The big challenge is part ii.

First you have $\int_{0}^{\frac{1}{2}}\frac{1}{\left(1+4x^{2}\right)^{n}}dx$
Next, integrate 1 which will give us $x$
As a result, your integration should look like this.
$\left[\frac{x}{\left(1+4x^{2}\right)^{n}}\right]_{0}^{\frac{1}{2}}+\int_{0}^{\frac{1}{2}}\frac{8nx^{2}}{\left(1+4x^{2}\right)^{n+1}}dx$
$\frac{1}{2^{n+1}}+\int_{0}^{\frac{1}{2}}\frac{8nx^{2}}{\left(1+4x^{2}\right)^{n+1}}dx$
$\frac{1}{2^{n+1}}+\int_{0}^{\frac{1}{2}}\frac{1}{\left(1+4x^{2}\right)^{n}}\times{\frac{8nx^{2}}{1+4x^{2}}dx$
$\frac{1}{2^{n+1}}+\int_{0}^{\frac{1}{2}}\frac{1}{\left(1+4x^{2}\right)^{n}}\times{2n\times{\frac{4x^{2}}{1+4x^{2}}}dx$
$\frac{1}{2^{n+1}}+2n\int_{0}^{\frac{1}{2}}\frac{1}{\left(1+4x^{2}\right)^{n}}\times{\frac{4x^{2}}{1+4x^{2}}dx$
$\frac{1}{2^{n+1}}+2n\int_{0}^{\frac{1}{2}}\frac{1}{\left(1+4x^{2}\right)^{n}}\times{\left(1-\frac{1}{1+4x^{2}}\right)}dx$
$I_{n}=\frac{1}{2^{n+1}}+2nI_{n}-2nI_{n+1}$
$I_{n}\left(1-2n\right)=\frac{1}{2^{n+1}}-2nI_{n+1}$
$I_{n}=\frac{1}{2^{n+1}\left(1-2n\right)}+\frac{-2nI_{n+1}}{-\left(2n-1\right)}$
$I_{n}=\frac{1}{2^{n+1}\left(1-2n\right)}+\frac{2nI_{n+1}}{2n-1}$

one more q (1 Viewer)

SCREW L+S

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5uckerberg

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