Evaluate the integral. Integral from 0 to pi/4 of 7*sec(theta)*tan(theta) d(theta).

Question:

Evaluate the integral.

{eq}\displaystyle \int_{0}^{\frac{\pi}{4}} 7 \sec(\theta) \tan(\theta) \, \mathrm{d}\theta {/eq}

Answer and Explanation:

For this particular definite integral, define {eq}u {/eq} as follows:

{eq}u = \sec \theta {/eq}

Then:

{eq}\mathrm{d}u = \sec \theta \ \tan \theta \ \mathrm{d}\theta {/eq}

Plugging in {eq}\mathrm{d}u = \sec \theta \ \tan \theta \ \mathrm{d}\theta {/eq}:

{eq}\begin{align*} \displaystyle \int_{0}^{\frac{\pi}{4}} 7 \sec(\theta) \tan(\theta) \, \mathrm{d}\theta & =\displaystyle 7\int_{0}^{\frac{\pi}{4}} \sec(\theta) \tan(\theta) \, \mathrm{d}\theta\\ & =\displaystyle 7\int_{0}^{\frac{\pi}{4}} \mathrm{d}u \;\;\; \left[ \mathrm{ U-Substitution }\right]\\ & =7\left[ u \right]_{0}^{\frac{\pi}{4}} \\ & = 7\left[ \sec \theta \right]_{0}^{\frac{\pi}{4}} \;\;\; \left[ \mathrm{ Substitute \ back \ }u= \sec \theta\right]\\ & = 7 \left(\sec\left( \frac{\pi}{4}\right) - \sec 0 \right) \\ & = 7 \left(\sqrt{2}- 1 \right) \\ \implies \displaystyle \int_{0}^{\frac{\pi}{4}} 7 \sec(\theta) \tan(\theta) \, \mathrm{d}\theta& = 7 \sqrt{2}-7\\ \end{align*} {/eq}


Learn more about this topic:

How to Solve Integrals Using Substitution

from Math 104: Calculus

Chapter 13 / Lesson 5
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