Mathematics

# $\displaystyle\int \frac{x}{\sqrt{\left ( 4-x^{4} \right )}}dx$

$\displaystyle \frac{1}{2}\sin ^{-1}\left ( \frac{1}{2}x^{2} \right ).$

##### SOLUTION
Let $\displaystyle I=\int \frac { x }{ \sqrt { \left( 4-x^{ 4 } \right) } } dx$

Put $\displaystyle x^{ 2 }=t\Rightarrow 2xdx=dt$
Therefore
$\displaystyle I=\frac { 1 }{ 2 } \int \frac { dt }{ \sqrt { \left( 2^{ 2 }-t^{ 2 } \right) } } =\frac { 1 }{ 2 } \sin ^{ -1 } \frac { t }{ 2 } =\frac { 1 }{ 2 } \sin ^{ -1 } \frac { { x }^{ 2 } }{ 2 }$
Hence, option 'C' is correct.

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

#### Realted Questions

Q1 Single Correct Medium
$\displaystyle \int \dfrac{x^4}{1+x^2}dx =$
• A. $\dfrac{x^2}{2}- tan^{-1} x+c$
• B. $x - tan^{-1} x+c$
• C. None of these
• D. $\dfrac{x^3}{3}- x+ tan^{-1} x+c$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Hard
Evaluate: $\displaystyle{\int \dfrac{1}{\sin x + \sec x}}\ dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Hard
$Find\,\displaystyle \int {\frac{{\sin x}}{{\sin \,4x}}dx}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Solve: $I=\int _{ o }^{ \pi }{ \dfrac { xdx }{ 1+\sin { x } } }$

The average value of a function f(x) over the interval, [a,b] is the number $\displaystyle \mu =\frac{1}{b-a}\int_{a}^{b}f\left ( x \right )dx$
The square root $\displaystyle \left \{ \frac{1}{b-a}\int_{a}^{b}\left [ f\left ( x \right ) \right ]^{2}dx \right \}^{1/2}$ is called the root mean square of f on [a, b]. The average value of $\displaystyle \mu$ is attained id f is continuous on [a, b].