What is \(\mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \frac{{{\rm{d\theta }}}}{{1 + \cos {\rm{\theta }}}}{\rm{\;}}\) equal to?

Concept:

• \(\smallint {\sec ^2}{\rm{x\;dx}} = {\rm{tan\;x}} + {\rm{c}}\)

Calculation:

Let I = \(\mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \frac{{{\rm{d\theta }}}}{{1 + \cos {\rm{\theta }}}}\)

\( \Rightarrow {\rm{I}} = \mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \frac{{{\rm{d\theta }}}}{{2{{\cos }^2}\left( {\frac{{\rm{\theta }}}{2}} \right)}}\) 

\( \Rightarrow {\rm{I}} = \frac{1}{2}\mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} {\sec ^2}\left( {\frac{{\rm{\theta }}}{2}} \right){\rm{d\theta }}\) 

Let θ/2 = t

⇒ (dθ)/2 = dt

\( \Rightarrow {\rm{I}} = \mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{4}} {\sec ^2}{\rm{tdt}}\) 

\(\Rightarrow {\rm{I}} = {\rm{\;}}\left[ {\tan {\rm{t}}} \right]_0^{\frac{{\rm{\pi }}}{4}} = {\rm{\;}}\tan \frac{{\rm{\pi }}}{4} - {\rm{\;}}\tan 0 = 1 - 0 = 1\)