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

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

Its FREE, you're just one step away


Subjective Medium Published on 17th 09, 2020
Next Question
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84
Enroll Now For FREE

Realted Questions

Q1 Single Correct Hard
The value of $$\int \dfrac {dx}{\sqrt {8 + 3x - x^{2}}}$$ is equal to
  • A. $$\dfrac {2}{3}\sin^{-1}\left (\dfrac {2x - 3}{\sqrt {41}}\right ) + C$$
  • B. $$\dfrac {3}{2}\sin^{-1}\left (\dfrac {2x - 3}{\sqrt {41}}\right ) + C$$
  • C. $$\dfrac {1}{\sqrt {41}}\sin^{-1}\left (\dfrac {2x - 3}{\sqrt {41}}\right ) + C$$
  • D. $$\sin^{-1}\left (\dfrac {2x - 3}{\sqrt {41}}\right ) + C$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q2 Subjective Medium
$$\int \frac{x^{2}+x-6}{(x-2)(x-1)}dx=$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q3 Single Correct Hard
Evaluate: $$\displaystyle  \int \left [ \frac{1-\sqrt{x}}{1+\sqrt{x}} \right ]^{1/2}\frac{dx}{x}$$
  • A. $$\displaystyle 2\sin ^{-1}\sqrt{x}+2\log \left [ \frac{1+\sqrt{1-x}}{\sqrt{x}} \right ]$$
  • B. $$\displaystyle 2\cos ^{-1}\sqrt{x}+2\log \left [ \frac{1-\sqrt{1+x}}{\sqrt{x}} \right ]$$
  • C. $$\displaystyle 2\sin ^{-1}\sqrt{x}-2\log \left [ \frac{1-\sqrt{1+x}}{\sqrt{x}} \right ]$$
  • D. $$\displaystyle 2\cos ^{-1}\sqrt{x}-2\log \left [ \frac{1+\sqrt{1-x}}{\sqrt{x}} \right ]$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q4 Single Correct Hard
If $$\displaystyle f(x)=\int\frac{d{x}}{\sqrt[3]{(x+1)^{2}(x-1)^{4}}}=k\sqrt[3]{\frac{1+x}{1-x}}+c$$, then $$k$$ is equal to
  • A. $$\dfrac {2}{3}$$
  • B. $$\dfrac {1}{3}$$
  • C. $$\dfrac {1}{2}$$
  • D. $$\dfrac {3}{2}$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q5 Passage Medium
Let $$F: R\rightarrow R$$ be a thrice differential function. Suppose that $$F(1) = 0, F(3) = -4$$ and $$F'(x)<0$$ for all $$x\in\left(\dfrac{1}{2},3\right)$$. Let $$f(x) = xF(x)$$ for all $$x\in R$$.

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer