Question

Evaluate the definite integral \(\int\limits_{\rm{0}}^{\rm{\pi }} {{\rm{Se}}{{\rm{c}}^{\rm{2}}}\left( {\frac{{\rm{t}}}{{\rm{4}}}} \right){\rm{dt}}} \).

Short answer

The value of the definite integral \(\int\limits_0^\pi {{\rm{Se}}{{\rm{c}}^2}} \left( {\frac{t}{4}} \right)dt\)is \(4\).

Step-by-step solution

Given Information.

The given definite integral is \(\int\limits_0^\pi {{\rm{Se}}{{\rm{c}}^2}} \left( {\frac{t}{4}} \right)dt\).

Make assumptions.

Let \(\frac{t}{4} = u\) then

\(\begin{aligned}{l}\frac{{dt}}{4} &= du\\ \Rightarrow dt &= 4du\end{aligned}\)

Change the limits.

If \(t = 0\) then \(u = 0\) and if \(t = \pi \) then \(u = \frac{\pi }{4}\)

So, the limits will be \(0,\frac{\pi }{4}\).

Solve the given definite integral.

\(\int\limits_0^\pi {{\rm{Se}}{{\rm{c}}^2}} \left( {\frac{t}{4}} \right)dt = 4\int\limits_0^{{\pi \mathord{\left/

{\vphantom {\pi 4}} \right.

\kern-\nulldelimiterspace} 4}} {{\rm{Se}}{{\rm{c}}^2}} udu\)

\(\begin{aligned}{l} &= 4\left( {\tan u} \right)_0^{{\pi \mathord{\left/

{\vphantom {\pi 4}} \right.

\kern-\nulldelimiterspace} 4}}\\ &= 4\left( {\tan \frac{\pi }{4} - \tan 0} \right)\\ &= 4\left( {1 - 0} \right)\\ &= 4\end{aligned}\)