Mathematics

# $\int _{ 0 }^{ \pi }{ x\log { \sin { xdx } } }$ is equal to

$-\dfrac { \pi }{ 2 } \log { \dfrac { 1 }{ 2 } }$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Subjective Medium
Evaluate :

$\displaystyle\int _{0}^{\pi/2} x\cos{2x}dx$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\displaystyle \int\frac{\sin 2x}{\sin^{4}x+\cos^{4}x}dx=$
• A. $\tan^{-1}(\cos^{2}x)+c$
• B. $\tan^{-1}(\sin^{2}x)+c$
• C. $\tan^{-1}(\cot^{2}x)+c$
• D. $\tan^{-1}(\tan^{2}x)+c$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Hard
$\int\limits_0^{\frac{\pi }{2}} {\frac{{x\sin \,2x\,dx}}{{\cos 4x + \sin 4x}}}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Find :  $\displaystyle\int \dfrac{sin x - cos x}{\sqrt{1 + sin \, 2x}} dx , \, 0 < x < \dfrac{\pi}{2}$

Let $\displaystyle f\left ( x \right )=\frac{\sin 2x \cdot \sin \left ( \dfrac{\pi }{2}\cos x \right )}{2x-\pi }$