Mathematics

# Evaluate the following integral:$\displaystyle \int { \cfrac { 1 }{ \sqrt { { (2-x) }^{ 2 }-1 } } } dx\quad$

##### SOLUTION
$\displaystyle\int{\dfrac{dx}{\sqrt{{\left(2-x\right)}^{2}-1}}}$

Let $t=2-x$

$\Rightarrow\,dt=-dx$

$=-\displaystyle\int{\dfrac{dt}{\sqrt{{t}^{2}-1}}}$

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

Replace $x\rightarrow\,t$ and $a\rightarrow\,1$

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

$=-\log{\left|2-x+\sqrt{{\left(2-x\right)}^{2}-1}\right|}+c$ where $t=2-x$

$=-\log{\left|2-x+\sqrt{4+{x}^{2}-2x-1}\right|}+c$

$=-\log{\left|2-x+\sqrt{{x}^{2}-2x+3}\right|}+c$

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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 Medium
$\int \frac{x\frac{1}{4}}{1+x\frac{1}{2}}dx$ is equal to
• A. $4(\frac{x\frac{2}{4}}{3}-x\frac{1}{4} tan^{-1}(x\frac{1}{4}))$
• B. $4(\frac{x\frac{2}{4}}{3}-x\frac{1}{4}- tan^{-1}(x\frac{1}{4}))$
• C. none of these
• D. $4(\frac{x\frac{2}{4}}{3}-x\frac{1}{4}+ tan^{-1}(x\frac{1}{4}))$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
$\displaystyle \int_{0}^{1}\displaystyle \frac{x(2x^{2}+1)}{x^{8}+2x^{6}-x^{2}+1}dx=$
• A. $\displaystyle \frac{\pi }{3}$
• B. $\displaystyle \frac{2\pi }{3}$
• C. $\displaystyle \frac{\pi }{6}$
• D. $\displaystyle \frac{\pi }{2\sqrt{3}}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
$\int {x{{\sin }^{ - 1}}xdx}$=?
• A. $\frac{1}{4}{\sin ^{ - 1}}(x)(2x + 1) + \frac{{x\sqrt {1 - {x^2}} }}{4} + c$
• B. $\frac{1}{4}{\cos ^{ - 1}}(x)(2x - 1) + \frac{{x\sqrt {1 - {x^2}} }}{4} + c$
• C. $\frac{1}{4}{\sin ^{ - 1}}(x)(2x - 1) - \frac{{x\sqrt {1 - {x^2}} }}{4} + c$
• D. $\frac{1}{4}{\sin ^{ - 1}}(x)(2x^2 - 1) + \frac{{x\sqrt {1 - {x^2}} }}{4} + c$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium

lf $\displaystyle \int_{0}^{\pi}$ x.f(Sinx)dx $=\displaystyle \mathrm{k}\int_{0}^{\pi /2}$ f(Sinx)dx then the value of $\mathrm{k}$ is
• A. $\pi/2$
• B. $\pi/3$
• C.
• D. $\pi$

Evaluate $\int {{e^3}} dx$