Mathematics

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

##### SOLUTION

Consider the following question.

$I=\int_{0}^{\dfrac{\pi }{2}}{x\sin x\cos xdx}$

Using a Ilate rule, we get.

Therefore,

$=\dfrac{1}{2}\int_{0}^{\dfrac{\pi }{2}}{x\sin 2xdx}$

$=\dfrac{1}{2}\int_{0}^{\dfrac{\pi }{2}}{x\sin 2xdx}$

$=\dfrac{1}{2}\left( \dfrac{1}{2}x\left( -\cos 2x \right)+\dfrac{1}{4}\sin 2x \right)_{0}^{\dfrac{\pi }{2}}$

$=\dfrac{1}{2}\left( x\left( -\cos 2x \right)+\dfrac{1}{2}\sin 2x \right)_{0}^{\dfrac{\pi }{2}}$

$=\dfrac{1}{2}\left( \dfrac{\pi }{2}\left( -\cos 2\dfrac{\pi }{2} \right)+\dfrac{1}{2}\sin 2\dfrac{\pi }{2} \right)-\left( 0\left( -\cos 2\times 0 \right)+\dfrac{1}{2}\sin 2\times 0 \right)$

$=\dfrac{\pi }{4}$

Hence, this is the required answer.

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

#### Realted Questions

Q1 Single Correct Hard
Let $f$ be a positive function. Let $\displaystyle I_1=\int_{\displaystyle 1-k}^{\displaystyle k}{(x)f(x(1-x))dx}$; $\displaystyle I_2=\int_{\displaystyle 1-k}^{\displaystyle k}{f(x(1-x))dx}$, where $2k-1>0$. Then, $\displaystyle\frac{I_2}{I_1}$ is
• A. $k$
• B. $\displaystyle\frac{1}{2}$
• C. $1$
• D. $2$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Solve :
$I = \int \dfrac {e^x}{ e^{2x} - 3e^x + 1 } dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Hard
Evaluate:
$\int { \cfrac { dx }{ ({ x }^{ 2 }+2)({ x }^{ 2 }+4) } }$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$\displaystyle \int \tan ^{-1}\sqrt{\left ( \frac{1-x}{1+x} \right )}dx.$
• A. $\displaystyle \frac{1}{2}\left [ x\cos ^{-1}x-\sqrt{1-x^{2}} \right ]$
• B. $\displaystyle -\left [ x\cos ^{-1}x-\sqrt{1-x^{2}} \right ]$
• C. $\displaystyle -\frac{1}{2}\left [ \cos ^{-1}x-\sqrt{1-x^{2}} \right ]$
• D. $\displaystyle -\frac{1}{2}\left [ x\cos ^{-1}x-\sqrt{1-x^{2}} \right ]$

Evaluate  $\int\limits_{ - \pi }^\pi {\frac{{2x\left( {1 + \sin x} \right)}}{{1 + {{\cos }^2}x}}} dx$