Question

Evaluate the integral.

\(\int_0^{\pi /4} {\frac{{1 + {{\cos }^2}\theta }}{{{{\cos }^2}\theta }}} d\theta \).

Short answer

The solution of given integral\(\int_0^{\pi /4} {\frac{{1 + {{\cos }^2}\theta }}{{{{\cos }^2}\theta }}} d\theta \) is \(1 + \frac{\pi }{4}\).

Step-by-step solution

Definition of integration or Antiderivative

The function\({\rm{F}}\)is called as antiderivative of\({\rm{f}}\)on an interval\({\rm{I}}\)if\({\rm{F'(x) = f(x)}}\)

i.e., \(\int {{\rm{f(x)dx = F(x)}}} \)

Applying definition of integration

Using evaluation theorem writes given integral as-

\(\begin{aligned}{c}\int_0^{\pi /4} {\frac{{1 + {{\cos }^2}\theta }}{{{{\cos }^2}\theta }}} d\theta &= \int_0^{\pi /4} {\left( {\frac{1}{{{{\cos }^2}\theta }} + \frac{{{{\cos }^2}\theta }}{{{{\cos }^2}\theta }}} \right)} d\theta \\ &= \int_0^{\pi /4} {\left( {\frac{1}{{{{\cos }^2}\theta }} + 1} \right)} d\theta \\ &= \int_0^{\pi /4} {\frac{1}{{{{\cos }^2}\theta }}} d\theta + \int_0^{\pi /4} {1{\rm{ }}d} \theta \\ &= \int_0^{\pi /4} {{{\sec }^2}} \theta d\theta + \int_0^{\pi /4} {1{\rm{ }}d} \theta \\ &= (\tan \theta )_0^{\pi /4} + (\theta )_0^{\pi /4}\\ &= \tan \frac{\pi }{4} - \tan 0 + \frac{\pi }{4} - 0\\ &= 1 - 0 + \frac{\pi }{4} - 0\\ &= 1 + \frac{\pi }{4}\end{aligned}\)

Therefore, the solution of given integral\(\int_0^{\pi /4} {\frac{{1 + {{\cos }^2}\theta }}{{{{\cos }^2}\theta }}} d\theta \) is \(1 + \frac{\pi }{4}\).