Mathematics

# $\displaystyle\int^{\pi}_0\sqrt{|\cos x|-|\cos^3x|}dx$ is?

$\dfrac{4}{3}$

##### SOLUTION
Given : $\int _{ 0 }^{ \pi }{ \sqrt { \left| \cos { x } \right| -\left| \cos ^{ 3 }{ x } \right| dx } }$
$I=\int _{ 0 }^{ \pi }{ \sqrt { \left| \cos { x } \right| -\left| \cos ^{ 3 }{ x } \right| dx } } =\int _{ 0 }^{ \pi }{ \sqrt { \left| \cos { x } \right| \left| 1-\cos ^{ 2 }{ x } \right| dx } } \\ I=\int _{ 0 }^{ \pi }{ \sqrt { \left| \cos { x } \right| } \left| \sin { x } \right| dx } \\ I=\int _{ 0 }^{ \pi }{ \sqrt { \left| \cos { x } \right| } \left( \sin { x } \right) dx } \\ \sin { x } >0\quad (0<x<\pi )\\ \cos { x } =t\quad x\rightarrow 0\quad t\rightarrow 1\\ dt=-\sin { x } dx\quad x\rightarrow \pi \quad t\rightarrow -1\\ I=-\int _{ 1 }^{ -1 }{ \sqrt { \left| t \right| } dt } \\ =\int _{ -1 }^{ 1 }{ \sqrt { \left| t \right| } dt } \\ =\int _{ -1 }^{ 0 }{ \sqrt { -t } dt } +\int _{ 0 }^{ 1 }{ \sqrt { t } dt } \\ I=-\cfrac { 2 }{ 3 } \left| \left( -t \right) ^{ \cfrac { 3 }{ 2 } } \right| _{ -1 }^{ 0 }+\cfrac { 2 }{ 3 } \left| t^{ \cfrac { 3 }{ 2 } } \right| _{ 0 }^{ 1 }\\ I=-\cfrac { 2 }{ 3 } \left[ 0-1 \right] +\cfrac { 2 }{ 3 } (1-0)\\ =\cfrac { 4 }{ 3 }$
Hence the correct answer is $\cfrac { 4 }{ 3 }$

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Single Correct Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

#### Realted Questions

Q1 Single Correct Hard
Evaluate: $\int \displaystyle \frac{dx}{x^{3} (x^{3} + 1)^{1/3}}$
• A. $\displaystyle \frac{1}{2} \left ( 1 - \frac{1}{x^{3}} \right )^{1/3} + c$
• B. $\displaystyle \frac{1}{2} \left ( 1 - \frac{1}{x^{3}} \right )^{2/3} + c$
• C. $\displaystyle -\frac{1}{2} \left ( 1 + \frac{1}{x^{3}} \right )^{1/3} + c$
• D. $\displaystyle -\frac{1}{2} \left ( 1 + \frac{1}{x^{3}} \right )^{2/3} + c$

1 Verified Answer | Published on 17th 09, 2020

Q2 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.

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Evaluate $\int_{0}^{1} x \tan^{-1} x\ dx$.

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
The value of $\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\log {{\left( {1 + \frac{K}{n}} \right)}^{\frac{1}{n}}}}$ is
• A. ${\log _e}\left( {\frac{4}{e}} \right)$
• B. ${\log _e}4$
• C. None if these
• D. ${\log _e}\left( {\frac{e}{4}} \right)$

$\displaystyle \int_{0}^{\pi}\dfrac {xdx}{1+\sin x} = \pi$