Mathematics

# Solve $\displaystyle\int \dfrac{dx}{(x^2 + a^2)^{3/2}}$

##### SOLUTION
$\int \dfrac{dx}{(x^2 + a^2)^{3/2}}$
Let x = a tan y.
$dx = a sec^2 y dy$
$\int \dfrac{a \, sec^2 y \, dy}{(a^2 tan^2 y + a^2)^{3/2}}$
$= \int \dfrac{a \, sec^2 y}{(a^2 sec^2 y)^{3/2}}$
$= \int \dfrac{a \, sec^2 y}{a^3 sec^3 y}$
$= \dfrac{1}{a^2} \int cos y$
$\dfrac{1}{a^2} sin y + c$
$\dfrac{x}{a} = tan y$
$cot y = \dfrac{a}{x}$
$cosec y = \sqrt{1 + \dfrac{a^2}{x^2}}$
$= \sqrt{\dfrac{x^2 = a^2}{x}} = \dfrac{1}{a^2} \dfrac{x}{\sqrt{x^2 + a^2}} + c$
$sin y = \dfrac{x}{\sqrt{x^2 + a^2}}$

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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 Hard

lf $\displaystyle \int\frac{2\sin x+3\cos x}{3\sin x+4\cos x}dx=A$ Iog $|3\sin x+4 \cos x|+$ Bx $+c$, then $A=\ldots\ldots\ldots,\ B=\ldots\ldots\ldots.$,
• A. $-\displaystyle \frac{1}{25},\frac{18}{25}$
• B. $-\displaystyle \frac{1}{5}, -\displaystyle \frac{1}{5}$
• C. $\displaystyle \frac{1}{25},\frac{3}{25}$
• D. $\displaystyle \frac{1}{25},\ \displaystyle \frac{18}{25}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
Match the following with I, II, III
If $\displaystyle \frac{x^{2}-x+3}{x^{3}-1}=\frac{A}{(x-1)}+\frac{Bx+C}{(x^{2}+x+1)}$ then

I) $A=$             a)  0
II) $B=$            b)  1
III) $C=$          c)  -2
• A. a, b, c
• B. b, c, a
• C. a, c,b
• D. b, a, c

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
$\int_{}^{} {\dfrac{{{x^2}}}{{\sqrt {1 - x} }}dx = \int_{}^{} {\dfrac{{{{\left( {1 - u} \right)}^2}}}{{\sqrt u }}\left( { - du} \right)} }$
• A. $\frac{{ - 1}}{{15}}\log \left( {1 - x} \right)\left( {3{x^2} + 4x + 8} \right)$
• B. $\frac{{ - 2}}{{15}}\sqrt {1 - x} \left( {3{x^2} + 4x + 8} \right)$
• C. $None of these$
• D. $\frac{2}{3}{\left( {1 - x} \right)^{\frac{3}{2}}}\left( {3{x^2} + 4x + 5} \right)$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Prove that:
$\displaystyle \int \tan^{3}2x\sec 2x\ dx$

Solve $\int {x{{\sin }^2}xdx}$