Mathematics

# Evaluate the following integral:$\int { \cfrac { 1 }{ \sqrt { 8+3x-{ x }^{ 2 } } } } dx\quad$

##### SOLUTION
Consider $I=\displaystyle\int { \dfrac { 1 }{ \sqrt { 8+3x-{ x }^{ 2 } } } } dx$

$=\displaystyle\int { \dfrac { 1 }{ \sqrt { -\left( { x }^{ 2 }-3x-8 \right) } } } dx$

$=\displaystyle\int { \dfrac { 1 }{ \sqrt { -\left( { x }^{ 2 }-3x+\cfrac { 9 }{ 4 } -\dfrac { 9 }{ 4 } -8 \right) } } } dx$

$\Rightarrow I=\displaystyle\int { \dfrac { 1 }{ \sqrt { -\left\{ { \left( x-\cfrac { 3 }{ 2 } \right) }^{ 2 }-{ \left( \dfrac { \sqrt { 41 } }{ 2 } \right) }^{ 2 } \right\} } } } dx$

$=\displaystyle\int { \dfrac { 1 }{ \sqrt { { \left( \dfrac { \sqrt { 41 } }{ 2 } \right) }^{ 2 }-{ \left( x-\cfrac { 3 }{ 2 } \right) }^{ 2 } } } } dx$

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

Replace $a\rightarrow\dfrac{\sqrt{41}}{2}$ and $x\rightarrow\,2x-3$

$=\sin ^{ -1 }{ \left( \dfrac { 2x-3 }{ \sqrt { 41 } } \right) } +C$

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

#### Realted Questions

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The value of $\displaystyle \int_{8}^{15}\displaystyle \frac{dx}{\left ( x-3 \right )\sqrt{x+1}}$ is
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• B. $2\log \displaystyle \frac{5}{3}$
• C. $\sqrt{2}\log \left ( \sqrt{2}-1 \right )$
• D. $\displaystyle \frac{1}{2}\log \displaystyle \frac{5}{3}$

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Q5 Single Correct Hard
$\displaystyle \int_{-\sqrt{2}}^{\sqrt{2}}\frac{2x^{7}+3x^{6}-10x^{5}-7x^{3}-12x^{2}+x+1}{x^{2}+2}dx= \cdots$
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