Mathematics

# $\displaystyle\int \dfrac{dx}{\sqrt{10-10x-2x^2}}$=?

##### SOLUTION
We are given $I=\displaystyle\int \dfrac{dx}{\sqrt{10-10x-2x^{2}}}$
$=\displaystyle\int \dfrac{dx}{\sqrt{\dfrac{45}{2}-(2x^{2}+10x+25/2)}}$
$=\displaystyle\dfrac{dx}{\sqrt{\left(\dfrac{3\sqrt{5}}{\sqrt{2}}\right)^{2}-\left(\sqrt{2}x+\dfrac{5}{\sqrt{2}}\right)^{2}}}$
as we know, $\displaystyle\int \dfrac{dx}{\sqrt{a^{2}-x^{2}}}=\sin^{-1}\left(\dfrac{x}{a}\right)+C$
$I=\sin^{-1}\left[\dfrac{\sqrt{2}x+5/\sqrt{2}}{\dfrac{3\sqrt{5}}{\sqrt{2}}}\right]+C$
$=\sin^{-1}\left(\dfrac{2x+5}{3\sqrt{5}}\right)+C$
So,
$\displaystyle\int \dfrac{1}{\sqrt{10-10x-2x^{2}}}dx=\sin^{--1}\left(\dfrac{2x+5}{3\sqrt{5}}\right)+C$
Where $C$ is any arbitary constant.

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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