Mathematics

# Evaluate the following integral:$\displaystyle \int { \cfrac { 1 }{ \sqrt { 5-4x-2{ x }^{ 2 } } } } dx$

##### SOLUTION
$\displaystyle\int{\dfrac{dx}{\sqrt{5-4x-2{x}^{2}}}}$

Consider $5-4x-2{x}^{2}=-\left(2{x}^{2}+4x-5\right)$

$=-2\left({x}^{2}+2x-\dfrac{5}{2}\right)$

$=-2\left({x}^{2}+2x+1-1-\dfrac{5}{2}\right)$

$=-2\left({\left(x+1\right)}^{2}+\dfrac{-2-5}{2}\right)$

$=-2\left({\left(x+1\right)}^{2}+\dfrac{-7}{2}\right)$

$=2\left(\dfrac{7}{2}-{\left(x+1\right)}^{2}+\right)$

$=2\left({\left(\dfrac{7}{2}\right)}^{2}-{\left(x+1\right)}^{2}+\right)$

$\displaystyle\int{\dfrac{dx}{\sqrt{5-4x-2{x}^{2}}}}$

$=\displaystyle\int{\dfrac{dx}{\sqrt{2\left({\left(\dfrac{7}{2}\right)}^{2}-{\left(x+1\right)}^{2}+\right)}}}$

$=\dfrac{1}{\sqrt{2}}\displaystyle\int{\dfrac{dx}{\sqrt{\left({\left(\dfrac{7}{2}\right)}^{2}-{\left(x+1\right)}^{2}+\right)}}}$

We know that $\displaystyle\int{\dfrac{dx}{\sqrt{{a}^{2}-{x}^{2}}}}={\sin}^{-1}{\dfrac{x}{a}}+c$

Replace $x\rightarrow\,x\rightarrow\,x+1$ and $a\rightarrow\,\sqrt{\dfrac{7}{2}}$

$=\dfrac{1}{\sqrt{2}}{\sin}^{-1}{\dfrac{x+1}{\sqrt{\dfrac{7}{2}}}}+c$

$=\dfrac{1}{\sqrt{2}}{\sin}^{-1}{\sqrt{\dfrac{2}{7}}\left(x+1\right)}+c$

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