Mathematics

# Find the integral$\int {\dfrac{{dx}}{{\sqrt {9x - 4{x^2}} }}}$

##### SOLUTION

Consider the given integral.

$I=\int{\dfrac{dx}{\sqrt{9x-4{{x}^{2}}}}}$

$I=\int{\dfrac{dx}{\sqrt{9x-4{{x}^{2}}+\dfrac{81}{16}-\dfrac{81}{16}}}}$

$I=\int{\dfrac{dx}{\sqrt{{{\left( \dfrac{9}{4} \right)}^{2}}-{{\left( 2x-\dfrac{9}{4} \right)}^{2}}}}}$

Put

$t=2x-\dfrac{9}{4}$

$\dfrac{dt}{2}=dx$

Therefore,

$I=\dfrac{1}{2}\int{\dfrac{dt}{\sqrt{{{\left( \dfrac{9}{4} \right)}^{2}}-{{t}^{2}}}}}$

$I=\dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{t}{\dfrac{9}{4}} \right)+C$

Put the value of t in above expression, we get

$I=\dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{4\left( 2x-\dfrac{9}{4} \right)}{9} \right)+C$

$I=\dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{\left( 8x-9 \right)}{9} \right)+C$

Hence, the value of this integral is $\dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{\left( 8x-9 \right)}{9} \right)+C$.

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

#### Realted Questions

Q1 Single Correct Medium
$\displaystyle \int _{ 0 }^{ { \pi }/{ 8 } }{ { \cos }^{ 3 } } 4\theta d\theta$ is equal to:
• A. $\displaystyle \frac { 5 }{ 3 }$
• B. $\displaystyle \frac { 5 }{ 4 }$
• C. $\displaystyle \frac { 1 }{ 3 }$
• D. $\displaystyle \frac { 1 }{ 6 }$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
Evaluate the given integral.
$\displaystyle\int { \cfrac { \sin ^{ 2 }{ x } }{ \cos ^{ 4 }{ x } } } dx$
• A. $\cfrac { 1 }{ 3 } \tan ^{ 2 }{ x } +C\quad \quad$
• B. $\cfrac { 1 }{ 2 } \tan ^{ 2 }{ x } +C\quad$
• C. None of these
• D. $\cfrac { 1 }{ 3 } \tan ^{ 3 }{ x } +C\quad$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium

$\displaystyle \int e^{2x}(\frac{\cot2x-1}{\cos x\sin x})dx=$
• A. 2 $e^{2x}\csc 2x+c$
• B. $e^{2x}\cot 2x+c$
• C. $e^{2x}cosec\ 2x+c$
• D. $-e^{2x}cosec\ 2x+c$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
Evaluate $\displaystyle\int {\dfrac{{{x^2}}}{{{x^4} + {x^2} - 2}}} \,dx$

Let $\displaystyle f\left ( x \right )=\frac{\sin 2x \cdot \sin \left ( \dfrac{\pi }{2}\cos x \right )}{2x-\pi }$