Mathematics

# The value of the definite integral$\overset { { a }_{ 1 } }{ \underset { { a }_{ 2 } }{ \int { \frac { d\theta }{ 1+tan\theta } } } } =\frac { 501\pi }{ K }$ where $\ a _{ 2 }=\quad \frac { 1003\pi }{ 2008 }$ and ${ \ a }_{ 1 }=\frac { \pi }{ 2008 }$ The value of K equalls

2008

##### SOLUTION
$\begin{array}{l} \int _{ { a_{ 2 } } }^{ { a_{ 1 } } }{ \frac { { d\theta } }{ { 1+\tan \theta } } } \\ I=\int _{ \frac { \pi }{ { 2008 } } }^{ \frac { { 1003\pi } }{ { 2008 } } }{ \frac { { d\theta } }{ { 1+\tan \theta } } } \\ I=\int _{ \frac { \pi }{ { 2008 } } }^{ \frac { { 1003\pi } }{ { 2008 } } }{ \frac { { \tan \theta } }{ { 1+\tan \theta } } }d\theta \\ 2I=\int _{ \frac { \pi }{ { 2008 } } }^{ \frac { { 1003\pi } }{ { 2008 } } }{ d\theta } \\ 2I=\frac { { 1002\pi } }{ { 2008 } } \\ I=\frac { { 501 } }{ { 2008 } } \pi \\ k=2008 \end{array}$
Option $D$ is correct answer.

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

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