Mathematics

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

$\displaystyle 2\cos ^{-1}\sqrt{x}-2\log \left [ \frac{1+\sqrt{1-x}}{\sqrt{x}} \right ]$

##### SOLUTION
Let $\displaystyle I=\int \left[ \frac { 1-\sqrt { x } }{ 1+\sqrt { x } } \right] ^{ 1/2 }\frac { dx }{ x }$

Substitute $\displaystyle x=\cos ^{ 2 } 2\theta ,\frac { 1-\cos 2\theta }{ 1+\cos 2\theta } =\tan ^{ 2 } \theta \Rightarrow dx=-4\cos .\sin 2\theta d\theta$

$\displaystyle I=-4\int \frac { \sin \theta }{ \cos \theta } .\frac { \cos 2\theta \left( 2\sin \theta .\cos \theta \right) }{ \cos ^{ 2 } 2\theta } d\theta =-4\int \frac { 2\sin ^{ 2 } \theta }{ \cos 2\theta } d\theta$

$\displaystyle =-4\int \frac { \left( 1-\cos 2\theta \right) }{ \cos 2\theta } d\theta =-4\int \left( \sec 2\theta -1 \right) d\theta =-4\left[ \frac { 1 }{ 2 } \log \left( \sec 2\theta +\tan 2\theta \right) -\theta \right]$

$\displaystyle I=4\theta -2\log \left[ \sec 2\theta +\sqrt { \sec ^{ 2 } 2\theta -1 } \right]$

Now put $\displaystyle \cos 2\theta =\sqrt { x } \Rightarrow \theta =\frac { 1 }{ 2 } \sec ^{ -1 } \frac { 1 }{ \sqrt { x } } =\frac { 1 }{ 2 } \cos ^{ -1 } \sqrt { x }$

$\displaystyle I=2\cos ^{ -1 } \sqrt { x } -2\log \left[ \frac { 1 }{ \sqrt { x } } +\sqrt { \frac { 1 }{ x } -1 } \right] =2\cos ^{ -1 } \sqrt { x } -2\log \left[ \frac { 1+\sqrt { 1-x } }{ \sqrt { x } } \right]$

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Single Correct Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 114

#### Realted Questions

Q1 Single Correct Medium
$\displaystyle \int\frac{e^{x}}{(2e^{x}-3)^{2}}dx=$
• A. $\displaystyle \frac{1}{2e^{x}-3}+c$
• B. $\displaystyle \frac{1}{2(2e^{x}-3)^{2}}+c$
• C. $\displaystyle \frac{1}{(2e^{x}-3)^{2}}+c$
• D. $\displaystyle \frac{-1}{2(2e^{x}-3)}+c$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
$\displaystyle\int^{1}_0\sin^{-1}\left(\dfrac{2x}{1+x^2}\right)dx$.

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
Evaluate: $\displaystyle \int_{0}^{1}\frac{\tan^{-1}x}{1+x^{2}}dx$
• A. $\displaystyle \frac{\pi^{2}}{4}$
• B. $\displaystyle \frac{\pi^{2}}{18}$
• C. $\displaystyle \frac{\pi^{2}}{8}-1$
• D. $\displaystyle \frac{\pi^{2}}{32}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$\lim_{x\rightarrow \infty }\left [ \left ( 1+\frac{1}{n^{2}}^{\frac{2}{n}^{2}} \right )\left ( 1+\frac{2^{2}}{n^{2}}^{\frac{4}{n^{2}}} \right )......\left ( 1+\frac{n^{2}}{n^{2}}^{\frac{2n}{n^{2}}} \right ) \right ]equals$
• A. $\frac{4}{e}$
• B. $\frac{6}{e}$
• C. $\frac{8}{e}$
• D. $\frac{2}{e}$

1 Verified Answer | Published on 17th 09, 2020

Q5 Passage Hard
In calculating a number of integrals we had to use the method of integration by parts several times in succession.
The result could be obtained more rapidly and in a more concise form by using the so-called generalized formula for integration by parts
$\displaystyle \int u\left ( x \right )v\left ( x \right )dx=u\left ( x \right )v_{1}-u'\left ( x \right )v_{2}\left ( x \right )+u''\left ( x \right )v_{3}\left ( x \right )+...+\left ( -1 \right )^{n-1}u^{n-1}\left ( x \right )V_{n}\left ( x \right ) \\ -\left ( -1 \right )^{n-1}\int u^{n}\left ( x \right )V_{n}\left ( x \right )dx$
where  $\displaystyle v_{1}\left ( x \right )=\int v\left ( x \right )dx,v_{2}\left ( x \right )=\int v_{1}\left ( x \right )dx ..., v_{n}\left ( x \right )= \int v_{n-1}\left ( x \right )dx$
Of course, we assume that all derivatives and integrals appearing in this formula exist. The use of the generalized formula for integration by parts is especially useful when  calculating $\displaystyle \int P_{n}\left ( x \right )Q\left ( x \right )dx,$ where $\displaystyle P_{n}\left ( x \right )$ is polynomial of degree n and the factor Q(x) is such that it can be integrated successively n+1 times.