Mathematics

# Let $f\left( x \right) =\frac { \sin { x } }{ x }$, then $\int _{ 0 }^{ \frac { \pi }{ 2 } }{ f\left( x \right) f\left( \frac { \pi }{ 2 } -x \right) } dx=$

$\frac { { 2 } }{ \pi } \int _{ 0 }^{ \pi }{ f\left( x \right)dx }$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

#### Realted Questions

Q1 Single Correct Medium
Match the following with I, II, III
If $\displaystyle \frac{x^{2}-x+3}{x^{3}-1}=\frac{A}{(x-1)}+\frac{Bx+C}{(x^{2}+x+1)}$ then

I) $A=$             a)  0
II) $B=$            b)  1
III) $C=$          c)  -2
• A. a, b, c
• B. b, c, a
• C. a, c,b
• D. b, a, c

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\displaystyle\int \tan ^{m}x\sec ^{2}xdx$
• A. $\displaystyle \frac{\tan ^{m}x}{m}.$
• B. $\displaystyle \frac{\sec ^{m+1}x}{m+1}.$
• C. $\displaystyle \frac{\tan ^{m+1}x}{m}.$
• D. $\displaystyle \frac{\tan ^{m+1}x}{m+1}.$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
The value of $\displaystyle\int { \tan ^{ 3 }{ 2x } \sec { 2x } } dx$ is equal to :
• A. $\dfrac { 1 }{ 3 } \sec ^{ 3 }{ 2x } -\dfrac { 1 }{ 2 } \sec { 2x } +C$
• B. $-\dfrac { 1 }{ 6 } \sec ^{ 3 }{ 2x } -\dfrac { 1 }{ 2 } \sec { 2x } +C$
• C. $\dfrac { 1 }{ 3 } \sec ^{ 3 }{ 2x } +\dfrac { 1 }{ 2 } \sec { 2x } +C$
• D. $\dfrac { 1 }{ 6 } \sec ^{ 3 }{ 2x } -\dfrac { 1 }{ 2 } \sec { 2x } +C$

1 Verified Answer | Published on 17th 09, 2020

Q4 Multiple Correct Hard
The value of $\left( \displaystyle \int _{ 0 }^{ \sin ^{ 2 } x } \sin ^{ -1 } \sqrt { t } dt \right) + \left( \displaystyle \int _{ 0 }^{ \cos ^{ 2 } x } \cos ^{ -1 } \sqrt { t } \, dt \right)$ is
• A. $\pi /4$
• B. none of these
• C. $\pi /2$
• D. $1$

The value of $\displaystyle\int{{e}^{\ln{\sqrt{x}}}dx}$ is