. In this question you will adapt two different ways to compute ſsec® de in order compute ſcsc0 d0. Adapting a solution you've seen to solve a new problem is an important skill in mathematics; it means that instead of just solving one problem, you can solve many different problems! (1) See p. 483 of the textbook for the most common strategy for integrating sece. Adapt this to compute Scsc0 d0.
Percentage
A percentage is a number indicated as a fraction of 100. It is a dimensionless number often expressed using the symbol %.
Algebraic Expressions
In mathematics, an algebraic expression consists of constant(s), variable(s), and mathematical operators. It is made up of terms.
Numbers
Numbers are some measures used for counting. They can be compared one with another to know its position in the number line and determine which one is greater or lesser than the other.
Subtraction
Before we begin to understand the subtraction of algebraic expressions, we need to list out a few things that form the basis of algebra.
Addition
Before we begin to understand the addition of algebraic expressions, we need to list out a few things that form the basis of algebra.
Pls Help!

![Isaac Barrow (1630–1677) in his Geometrical Lectures (Lect. XII, App. I) gave the first
"intelligible" proof of the result, but it was couched in the geometric idiom of the day. It is
especially noteworthy in that it is the earliest use of partial fractions in integration. Thus we
reproduce it here in modern garb:
1
de
cos e
O do =
cos e
de
-) Cos? 0
cos 0
de
=J1- sin?0
cos e
do
-J (1- sin0)(1+ sin 0)
cos e
1- sin0
cos 0
1+sine
+
do
-In|1-sine|+In|1+sin |]+ c
=
1+ sin0
-n1-sine
1+ sin0 1+sin0
In
1-sin0 1+sin 0
+c
|(1+sin0)²
-n1-sin*o
-In
+c
|(1+sin0)?
In
+c
=
(cos8)?
1+sin0
+ c
= In cose
=In|sec0+ tan 0|+c.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fde24d8a6-50dc-4080-850d-249835a4e10e%2Ffe6cdb3e-07c4-43ac-8099-764e33d75497%2F6tr2316_processed.png&w=3840&q=75)
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Solved in 2 steps with 1 images
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