# Evaluate the definite integral: Integral of (sin t)/(cos^2 t) dt from 0 to pi/4.

## Question:

Evaluate the definite integral: {eq}\int_{0}^{\frac{\pi }{4}} \frac{\sin t}{\cos^2 t} \mathrm{d}t {/eq}.

## Definite Integrals:

If {eq}\;\;\int_{a}^{b} f(x) \mathrm{d}x\;=\;[F(x)]_{a}^{b} \;=\; F(b)\;-\;F(a) {/eq}

- {eq}\int \sec x \tan x \mathrm{d}x\;=\;\sec x {/eq}

## Answer and Explanation:

{eq}\int_{0}^{\frac{\pi }{4}} \frac{\sin t}{\cos^2 t} \mathrm{d}t\;=\;\int_{0}^{\frac{\pi }{4}}\sec t \tan t \mathrm{d}t\;=\;[\sec t]_{0}^{\frac{\pi}{4}} {/eq}

{eq}\int_{0}^{\frac{\pi }{4}} \frac{\sin t}{\cos^2 t} \mathrm{d}t\;=\;[\sec t]_{0}^{\frac{\pi}{4}}\;=\; \sec( \frac{\pi}{4})\;-\;sec 0\;=\;1.414\;-\;1\;=\;0.414 {/eq}

Therefore,

{eq}\int_{0}^{\frac{\pi }{4}} \frac{\sin t}{\cos^2 t} \mathrm{d}t\;=\;0.414 {/eq}

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