   Chapter 3.9, Problem 33E

Chapter
Section
Textbook Problem

# 23-42 Find f. f ′ ( t ) = sec t ( sec t + tan t ) ,    − π / 2 < t < π / 2 , f ( π / 4 ) = − 1

To determine

To find:

The function ft

Explanation

1) Concept:

If F is an antiderivative of f on an interval I, then the most general antiderivative of f on I is Fx+c where c is an arbitrary constant.

Definition:

A function F  is called an antiderivative of f on an interval I if F'x=fx  f for all x in I.

2) Given:

f't=sectsect+tant, -π2<t<π2 ,fπ4=-1

3) Calculations:

We have f't= sectsect+tant

f't=sec2t+sect×tant

The general antiderivative of f't using rules of antiderivative is,

ft=tant+sect+C, Where C is the arbitrary constant

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