Mathematics

# $\displaystyle \int{{{x^3}\sin \left( {{{\tan }^{ - 1}}{x^4}} \right)} \over {1 + {x^8}}}dx$

##### SOLUTION
Solution : -

$I = \displaystyle \int{{{x^3}\sin \left( {{{\tan }^{ - 1}}{x^4}} \right)} \over {1 + {x^8}}}dx$

Let $\displaystyle tan^{-1}x^{4} = t$

$\displaystyle \Rightarrow \frac{4x^{3}}{1+x^{8}}dx = dt$
$\displaystyle \Rightarrow \frac{x^{3}}{1+x^{8}}dx = \frac{dt}{4}$
$\displaystyle \Rightarrow I = \int \frac{sin(t).dt}{4}$

$\displaystyle = \frac{-1}{4}[cost]+c$
$= \dfrac{-cost}{4}+c$

Now $\displaystyle t = tan^{-1} x^{4}$

$\displaystyle \therefore I = \frac{-cos(tan^{-1}x^{4})}{4}+c$

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

#### Realted Questions

Q1 Single Correct Hard
The value of $\int \dfrac {dx}{\sqrt {8 + 3x - x^{2}}}$ is equal to
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1 Verified Answer | Published on 17th 09, 2020

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$\int \frac{x^{2}+x-6}{(x-2)(x-1)}dx=$

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Q3 Single Correct Hard
Evaluate: $\displaystyle \int \left [ \frac{1-\sqrt{x}}{1+\sqrt{x}} \right ]^{1/2}\frac{dx}{x}$
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