Mathematics

# If $\int _{ 0 }^{ \pi }{ \frac { \sin { x } }{ { x }^{ 2 } } } dx,\quad then\quad \int _{ 0 }^{ { \pi }/{ 2 } }{ \frac { \cos { 2x } }{ x } } dx$ is equal to :

$1+A$

##### SOLUTION
Given,

$A=\int _0^{\pi }\:\:\dfrac{sinx}{x^2} dx$

integration by parts $u=\sin \left(x\right),\:v'=\dfrac{1}{x^2}$

$=\left[\dfrac{-\sin \left(x\right)-x\cdot \int \:-\frac{\cos \left(x\right)}{x}dx}{x}\right]^{\pi }_0$

$=\left[\dfrac{-\sin \left(x\right)-x\left(-\text{Ci}\left(x\right)\right)}{x}\right]^{\pi }_0$

$=\left[\dfrac{-\sin \left(x\right)+x\text{Ci}\left(x\right)}{x}\right]^{\pi }_0$

$=\text{Ci}\left(\pi \right)+1-\text{Ci}\left(0\right)$

$A=\text{Ci}\left(\pi \right)+1-\text{Ci}\left(0\right)$

Now,

$\int _0^{\frac{\pi }{2}}\dfrac{\cos \left(2x\right)}{x}dx$

substitute $u=2x$

$=\int _0^{\pi }\dfrac{\cos \left(u\right)}{u}du$

$=\left[\text{Ci}\left(u\right)\right]^{\pi }_0$

$=\text{Ci}\left(\pi \right)-\text{Ci}\left(0\right)$

$\therefore \int _0^{\frac{\pi }{2}}\dfrac{\cos \left(2x\right)}{x}dx=A+1$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Single Correct Medium
The value of $\displaystyle \int _{ 0 }^{ \pi }{ \sin ^{ 50 }{ x } \cos ^{ 49 }{ x } dx }$ is
• A. $\dfrac{\pi}{4}$
• B. $\dfrac{\pi}{2}$
• C. $1$
• D. $0$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
If $\displaystyle \int x^{5}(1+x^{3})^{2/3}dx=A(1+x^{3})^{8/3}+B(1+x^{3})^{5/3}+c$, then
• A. $A=\dfrac{1}{8},B=-\dfrac{1}{5}$
• B. $A=-\dfrac{1}{8},B=\dfrac{1}{5}$
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• D. $A=\dfrac{1}{4},B=\dfrac{1}{5}$

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$\displaystyle \int\frac{x^{2}}{x^{6}+2x^{3}-3}dx=$
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1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$\displaystyle \int {\cos 2\theta \,\log \left( {\frac{{\cos \theta + \sin \theta }}{{\cos \theta - \sin \theta }}} \right)} d\theta$ is equal to.
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Evaluate $\displaystyle \int_0^{10\pi}|\sin\,x|dx$