Mathematics

# If $\displaystyle\int _{ 0 }^{ \pi }{ xf\left( \sin ^{ 2 }{ x } +\sec ^{ 2 }{ x } \right) dx } =k \displaystyle\int _{ 0 }^{ { \pi }/{ 2 } }{ f\left( \sin ^{ 2 }{ x } +\sec ^{ 2 }{ x } \right) dx }$, then the value of $k$ is

$\pi$

##### SOLUTION
We have, $\displaystyle\int _{ 0 }^{ \pi }{ xf\left( \sin ^{ 2 }{ x } +\sec ^{ 2 }{ x } \right) dx } =k \displaystyle\int _{ 0 }^{ { \pi }/{ 2 } }{ f\left( \sin ^{ 2 }{ x } +\sec ^{ 2 }{ x } \right) dx }$
Let $I=\displaystyle\int _{ 0 }^{ \pi }{ xf\left( \sin ^{ 2 }{ x } +\sec ^{ 2 }{ x } \right) dx }$               .....(i)
$=\displaystyle\int _{ 0 }^{ \pi }{ \left( \pi -x \right) f\left( \sin ^{ 2 }{ \left( \pi -x \right) } +\sec ^{ 2 }{ \left( \pi -x \right) } \right) dx }$
$=\displaystyle\int _{ 0 }^{ \pi }{ \left( \pi -x \right) f\left( \sin ^{ 2 }{ x } +\sec ^{ 2 }{ x } \right) dx }$                 ....(ii)
On adding equations (i) and (ii), we get
$2I=\pi \displaystyle\int _{ 0 }^{ \pi }{ f\left( \sin ^{ 2 }{ x } +\sec ^{ 2 }{ x } \right) dx }$
$\Rightarrow 2I=2\pi \displaystyle\int _{ 0 }^{ { \pi }/{ 2 } }{ f\left( \sin ^{ 2 }{ x } +\sec ^{ 2 }{ x } \right) dx }$
$\Rightarrow I=\pi \displaystyle\int _{ 0 }^{ { \pi }/{ 2 } }{ f\left( \sin ^{ 2 }{ x } +\sec ^{ 2 }{ x } \right) dx }$
On comparing with given integral, we get $k=\pi$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

#### Realted Questions

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If $\displaystyle \int \frac{\cos{2x}+\sin 2x}{\left ( 2\cos x-\sin x \right )^{2}}dx=\frac{-A}{1725}x-\frac{2}{5}\log \left | 2\cos x-\sin x \right |+2\frac{1}{2-\tan x}+C$ then A is equal to

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
$\displaystyle \int_{e^{e^{e}}}^{e^{e^{e^{e}}}}\frac{dx}{xlnx\cdot ln\left ( lnx \right )\cdot ln\left ( ln\left ( lnx \right ) \right )}$ equals
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Q3 Single Correct Medium
$\displaystyle \int \sqrt{\left(\dfrac{1-\sqrt{x}}{1+\sqrt{x}}\right)}dx$ is equal to
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1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
If $b > a,$ and $\displaystyle I = \int _{a}^{b} \sqrt{\frac{ x-a}{b-x} }dx,$ then $I$ equals
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