Mathematics

# If $\displaystyle \int _{ 0 }^{ 1 }{ \frac { \sin { t } }{ 1+t } dt } =\alpha$, then the value of the integral $\displaystyle \int _{ 4\pi -2 }^{ 4\pi }{ \frac { \sin { t/2 } }{ 4\pi +2-t } dt }$ in terms of $\alpha$ is given by

$-\alpha$

##### SOLUTION
$\displaystyle \int _{ 4\pi -2 }^{ 4\pi }{ \frac { \sin { t/2 } }{ 4\pi +2-t } dt } =\frac { 1 }{ 2 } \int _{ 4\pi -2 }^{ 4\pi }{ \frac { \sin { t/2 } }{ 1+\left( 2\pi -\dfrac { t }{ 2 } \right) } dt }$

$\displaystyle =2.\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ \frac { \sin { \left( 2\pi -u \right) } }{ 1+u } du }$

$($ Substitute $\displaystyle 2\pi -\frac { t }{ 2 } =u\Rightarrow dt=-2du)$

$\displaystyle =\int _{ 0 }^{ 1 }{ \frac { \sin { u } }{ 1+u } du } =-\int _{ 0 }^{ 1 }{ \frac { \sin { t } }{ 1+t } dt } =-\alpha$

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Single Correct Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 114

#### Realted Questions

Q1 Subjective Medium
$\int_{0}^{5}x^{3} (25 - x^{2})^{7/2} dx$.

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
The value of $\int _{ }^{ }{ \cfrac { \log { x } }{ { \left( x+1 \right) }^{ 2 } } } dx$ is
• A. $\cfrac { \log { x } }{ x+1 } +\log { x } -\log { \left( x+1 \right) } +C$
• B. $\cfrac { \log { x } }{ x+1 } -\log { x } -\log { \left( x+1 \right) } + C$
• C. $\cfrac { -\log { x } }{ x+1 } -\log { x } -\log { \left( x+1 \right) } +C$
• D. $\cfrac { -\log { x } }{ x+1 } +\log { x } -\log { \left( x+1 \right) } +C$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
$\int \dfrac{1}{1-cos\dfrac{x}{2}}$ dx

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
If $I= \displaystyle \int_{-\pi /2}^{\pi /2}\displaystyle \frac{dx}{e^{\sin x}+1}$ then $I$ equals
• A. $3\pi /2$
• B. $\pi$
• C. $3\pi /4$
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Consider two differentiable functions $f(x), g(x)$ satisfying $\displaystyle 6\int f(x)g(x)dx=x^{6}+3x^{4}+3x^{2}+c$ & $\displaystyle 2 \int \frac {g(x)dx}{f(x)}=x^{2}+c$. where $\displaystyle f(x)>0 \forall x \in R$