Mathematics

# Solve$\displaystyle \int_{0}^{\pi /2}\left ( 2\log \sin x-\log \sin 2x \right )dx$

$-\dfrac {\pi}{2}\log 2$

##### SOLUTION
$\displaystyle I= \int_{0}^{\pi /2}\left ( 2\log \sin x-\log 2\sin x\cos x \right )dx$

$\displaystyle = \int_{0}^{\pi /2}\left [ 2\log \sin x-\left \{ \log 2+\log \sin x+\log \cos x \right \} \right ]dx$

$\displaystyle = \int_{0}^{\pi /2}\left [ \left ( \log \sin x-\log \cos x \right )-\log2 \right ]dx$

$\displaystyle = 0-\left [ x\log 2 \right ]_{0}^{\pi /2}= -\frac{\pi }{2}\log 2= \frac{\pi }{2}\log \frac{1}{2}$

$\because \int _{ 0 }^{ \pi /2 } \log \sin xdx=\int _{ 0 }^{ \pi /2 } \log \sin { \left( \dfrac { \pi }{ 2 } -x \right) } dx=\int _{ 0 }^{ \pi /2 } \log \cos xdx$ by integration property

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

#### Realted Questions

Q1 Subjective Medium
Evaluate:$\displaystyle \int \frac{(2x-1)}{(2x^{2}+2x+1)}dx$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
If $\displaystyle f\left ( \frac{1}{x} \right )+x^{2}f\left ( x \right )=0$ for $x> 0,$
and $\displaystyle I=\int_{1/x}^{x}f\left ( z \right )dz, \frac{1}{2}\leq x\leq 2$
then $\displaystyle I$ is?
• A. $\displaystyle f\left ( 2 \right )-f\left ( 1/2 \right )$
• B. $\displaystyle f\left ( 1/2 \right )-f\left ( 2 \right )$
• C. None of these
• D. $0$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
The value of integral $\displaystyle \int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \cfrac { \sin { x } -x\cos { x } }{ x(x+\sin { x } ) } } dx$ is
• A. $\log _{ e }{ \left\{ \cfrac { (\pi +3) }{ 2\left( 2\pi +3\sqrt { 3 } \right) } \right\} }$
• B. $\log _{ e }{ \left\{ \cfrac { \left( 2\pi +3\sqrt { 3 } \right) }{ 2(\pi +3) } \right\} }$
• C. $\log _{ e }{ \left\{ \cfrac { 2\left( 2\pi +3\sqrt { 3 } \right) }{ (\pi +3) } \right\} }$
• D. $\log _{ e }{ \left\{ \cfrac { 2(\pi +3) }{ \left( 2\pi +3\sqrt { 3 } \right) } \right\} }$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
Evaluate $\int \dfrac{1}{x^2+2x+2} dx$
• A. $\tan^{-1}(x+2)+C$
• B. $\tan^{-1}(x+3)+C$
• C. None of these
• D. $\tan^{-1}(x+1)+C$

$\int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx}$