Question

The value of the definite integral $\underset{{\theta }_1}{\overset{{\theta }_2}{\int }}\frac{d\theta }{1+\tan \theta }$ where ${\theta }_2=\frac{1004\pi }{2010}$ and ${\theta }_1=\frac{\pi }{2010}$ is equal to

• A. $\frac{1004\pi }{1005}$
• B. $\frac{1003\pi }{1005}$
• C. $\frac{1003\pi }{4020}$
• D. $\frac{1004\pi }{2010}$

Answer 1

Answer: (C)

$I=\underset{{\theta }_1}{\overset{{\theta }_2}{\int }}\frac{d\theta }{1+\tan \theta };{\theta }_1+{\theta }_2=\frac{\pi }{2}$
$I=\underset{{\theta }_1}{\overset{{\theta }_2}{\int }}\frac{d\theta }{1+\cot \theta }=\underset{{\theta }_1}{\overset{{\theta }_2}{\int }}\frac{\tan \theta d\theta }{1+\tan \theta }$ (using King)
$2I=\underset{{\theta }_1}{\overset{{\theta }_2}{\int }}d\theta ={\theta }_2-{\theta }_1=\frac{1003\pi }{2010}\Rightarrow I=\frac{1003\pi }{4020}$