Mathematics

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


ANSWER

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


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

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

Its FREE, you're just one step away


Single Correct Hard Published on 17th 09, 2020
Next Question
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 114
Enroll Now For FREE

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. $$\displaystyle \frac{-1}{2(2e^{x}-3)}+c$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

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

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
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. $$\displaystyle \frac{\pi^{2}}{32}$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
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. $$\frac{2}{e}$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
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.

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer