The value of the definite integral a 1 a 2 - d- 1-tan- - 501- K where a 2 - 1003- 2008 and a 1 - 2008 The value of K equalls

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Standard XII

Mathematics

Integration by Substitution

The value of ...

Question

The value of the definite integral
$_{1} = \frac{501 π}{K}$ where $a_{2} = \frac{1003 π}{2008}$ and $a_{1} = \frac{π}{2008}$ The value of K equalls

2007

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2006

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2009

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2008

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Solution

The correct option is D 2008
$\begin{matrix} \int_{a_{2}}^{a_{1}} \frac{d θ}{1 + tan θ} I = \int_{\frac{π}{2008}}^{\frac{1003 π}{2008}} \frac{d θ}{1 + tan θ} I = \int_{\frac{π}{2008}}^{\frac{1003 π}{2008}} \frac{tan θ}{1 + tan θ} d θ 2 I = \int_{\frac{π}{2008}}^{\frac{1003 π}{2008}} d θ 2 I = \frac{1002 π}{2008} I = \frac{501}{2008} π k = 2008 \end{matrix}$
Option $D$ is correct answer.

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The value of the definite integrala1∫dθ1+tanθa2=501πK where a2=1003π2008 and a1=π2008 The value of K equalls

The correct option is D 2008∫a1a2dθ1+tanθI=∫1003π2008π2008dθ1+tanθI=∫1003π2008π2008tanθ1+tanθdθ2I=∫1003π2008π2008dθ2I=1002π2008I=5012008πk=2008Option D is correct answer.

The value of the definite integral
$_{1} = \frac{501 π}{K}$ where $a_{2} = \frac{1003 π}{2008}$ and $a_{1} = \frac{π}{2008}$ The value of K equalls