Mathematics

# Evaluate the following integral:$\displaystyle \int { \cfrac { 1 }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 }{ x }^{ 2 } } } } dx$

##### SOLUTION
$\displaystyle\int{\dfrac{dx}{\sqrt{{a}^{2}+{b}^{2}{x}^{2}}}}$

$=\displaystyle\int{\dfrac{dx}{\sqrt{{a}^{2}+{\left(bx\right)}^{2}}}}$

Let $t=bx\Rightarrow\,dt=b\,dx$

$=\dfrac{1}{b}\displaystyle\int{\dfrac{dx}{\sqrt{{a}^{2}+{t}^{2}}}}$

We know that $\displaystyle\int{\dfrac{dx}{\sqrt{{a}^{2}+{x}^{2}}}}=\log{\left|x+\sqrt{{x}^{2}+{a}^{2}}\right|}+c$

Replace $x\rightarrow\,t$ and $a\rightarrow\,a$

$=\dfrac{1}{b}\log{\left|t+\sqrt{{t}^{2}+{a}^{2}}\right|}+c$

$=\dfrac{1}{b}\log{\left|bx+\sqrt{{b}^{2}{x}^{2}+{a}^{2}}\right|}+c$ where $t=bx$

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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
Solve $\displaystyle \int \dfrac{1}{2x^2 - 3} dx =$
• A. $\dfrac{1}{\sqrt 6} log \vert{\dfrac{\sqrt 2 x - \sqrt 3}{\sqrt 2 x + \sqrt 3}}\vert + c$
• B. $\dfrac{1}{2 \sqrt 6} log \vert{\dfrac{\sqrt 2 x + \sqrt 3}{\sqrt 2 x - \sqrt 3}}\vert + c$
• C. $\dfrac{1}{\sqrt 6} log \vert{\dfrac{\sqrt 2 x + \sqrt 3}{\sqrt 2 x - \sqrt 3}}\vert + c$
• D. $\dfrac{1}{2 \sqrt 6} log \vert{\dfrac{\sqrt 2 x - \sqrt 3}{\sqrt 2 x + \sqrt 3}}\vert + c$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Evaluate the definite integral:
$\displaystyle\int_{0}^{\pi/4}\sin^{3}2t\cos 2t\ dt$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
$\int \dfrac {1}{x^{2} (x^{4} + 1)^{3/4}}dx$ is equal to
• A. $\left (1 + \dfrac {1}{x^{4}}\right )^{1/4} + C$
• B. $(x^{4} + 1)^{1/4} + C$
• C. $\left (1 - \dfrac {1}{x^{4}}\right )^{1/4} + C$
• D. $-\left (1 + \dfrac {1}{x^{4}}\right )^{1/4} + C$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
If $\displaystyle \int_0^{\pi} \dfrac{x^2}{(1+\sin \,x)^2} dx = A$ the $\displaystyle \int_0^{\pi} \dfrac{2x^2cos^2(x/2)}{(1+\sin x)^2} dx=$ ?
• A. $A-\pi+ \pi^2$
• B. $A-\pi - \pi^2$
• C. $A+2\pi - \pi^2$
• D. $A+\pi - \pi^2$

$\displaystyle\int \dfrac{1}{e^x+e^{-x}}dx$.