Mathematics

# Evaluate: $\displaystyle \int_{0}^{1}\frac{x^{3}}{1+x^{8}} dx$

$\displaystyle \frac{\pi}{16}$

##### SOLUTION
Let, $z=x^4\implies dz=4x^3dx\implies x^3dx=\dfrac{dz}{4}$

When $x=0,z=0$ and when $x=1,z=1$

Hence,integration becomes:-

$\displaystyle\dfrac{1}{4}\int_{0}^{1} \dfrac{dz}{1+z^2}$

$=\dfrac{1}{4}\displaystyle\left[\tan^{-1}{z}\right]_{0}^{1}$

$=\dfrac{1}{4}\displaystyle\left[\dfrac{\pi}{4}-0\right]$

$=\dfrac{\pi}{16}$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 114

#### Realted Questions

Q1 Single Correct Hard
If $\displaystyle \int_{1}^{x}\dfrac{dt}{|t|\sqrt{t^{2}-1}}=\dfrac{\pi}{6}$, then $x$ can be equal to :
• A. $\sqrt{3}$
• B. $2$
• C. $\dfrac{4}{\sqrt{3}}$
• D. $\dfrac{2}{\sqrt{3}}$

1 Verified Answer | Published on 17th 09, 2020

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

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
Let $f$ be a positive function. Let
${ I }_{ 1 }=\int _{ 1-k }^{ k }{ xf\left\{ x(1-x) \right\} } dx$
${ I }_{ 2 }=\int _{ 1-k }^{ k }{ f\left\{ x(1-x) \right\} } dx$
where $2k-1>0$. Then $\cfrac { { I }_{ 1 } }{ { I }_{ 2 } }$
• A. $2$
• B. $k$
• C. $1$
• D. $\cfrac { 1 }{ 2 }$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
Integrate:
$\displaystyle \int{\dfrac{\sqrt{1-x^{2}}+\sqrt{1+x^{2}}}{\sqrt{1-x^{4}}}}dx=$

Consider two differentiable functions $f(x), g(x)$ satisfying $\displaystyle 6\int f(x)g(x)dx=x^{6}+3x^{4}+3x^{2}+c$ & $\displaystyle 2 \int \frac {g(x)dx}{f(x)}=x^{2}+c$. where $\displaystyle f(x)>0 \forall x \in R$