Mathematics

# Evaluate :$\displaystyle \int x^{3}\sqrt{1-x^{8}}dx$

##### SOLUTION
$I=\int { { x }^{ 3 }\sqrt { 1-{ x }^{ 8 } } dx } =\int { { x }^{ 3 }\sqrt { 1-{ ({ x }^{ 4 }) }^{ 2 } } dx }$
Putting ${ x }^{ 4 }=t$
Differentiating both sides
$4{ x }^{ 3 }dx=dt\\ dx=\cfrac { dt }{ 4{ x }^{ 3 } }$
$I=\cfrac { 1 }{ 4 } \int { \sqrt { 1-{ t }^{ 2 } } dt } =\cfrac { 1 }{ 4 } \{ [\cfrac { t }{ 2 } \sqrt { 1-{ t }^{ 2 } } +\cfrac { 1 }{ 2 } \sin ^{ -1 }{ t } ]+C\} \\ \because \int { \sqrt { { a }^{ 2 }-{ x }^{ 2 } } dx } =\cfrac { x }{ 2 } \sqrt { { a }^{ 2 }-{ x }^{ 2 } } +\cfrac { { a }^{ 2 } }{ 2 } \sin ^{ -1 }{ \cfrac { x }{ a } } +C$
Putting the value $t={ x }^{ 4 }$
$I=\cfrac { 1 }{ 8 } [{ x }^{ 4 }\sqrt { 1-{ x }^{ 8 } } +\sin ^{ -1 }{ { x }^{ 4 } } ]+C$

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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
The quadratic polynomial $p(x)$ has the following properties: $p(x)\ge 0$ for all real number, $p(1)=$ and $p(2)=2$ value of $p(0)+p(3)$ is equal to
• A. $9$
• B. $8$
• C. $None \ of \ these$
• D. $10$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Evaluate the integral   $\displaystyle \int_0^1\sin^{-1}\left (\frac {2x}{1+x^2}\right )dx$   using substitution.

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
Evaluate $\displaystyle \int_{}^{} {x\sqrt {\frac{{{a^2} - {x^2}}}{{{a^2} + {x^2}}}} dx }$
• A. $\displaystyle \frac{1}{2}{a^2}{\cos ^{ - 1}}\left( {\frac{{{x^2}}}{{{a^2}}}} \right) + \frac{1}{2}\sqrt {{a^4} + {x^4}} + C$
• B. $\displaystyle \frac{1}{2}{\sin ^{ - 1}}\left( {\frac{{{x^2}}}{{{a^2}}}} \right) + \sqrt {{a^4} + {x^4}} + C$
• C. $\displaystyle \frac{1}{2}{\cos ^{ - 1}}\left( {\frac{{{x^2}}}{{{a^2}}}} \right) + \frac{1}{2}\sqrt {{a^4} - {x^4}} + C$
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1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
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• C. $none\ of\ these$
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Evaluate $\displaystyle \int_0^{10\pi}|\sin\,x|dx$