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Found in: Page 289

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Evaluate the integral.$$\int_0^{\pi /4} {\frac{{1 + {{\cos }^2}\theta }}{{{{\cos }^2}\theta }}} d\theta$$.

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

See the step by step solution

## Step 1: 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., $$\int {{\rm{f(x)dx = F(x)}}}$$

## Step 2: 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 $$1 + \frac{\pi }{4}$$.