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Pillar of Autumn's question at Yahoo! Answers regarding an indefinite integral

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MarkFL

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Staff member
Feb 24, 2012
13,775
Here is the question:

What is this indefinite integral?


at 0 (angle)

(sin0 - 1)/cos^2(0) d0

Can you please list the steps
I have posted a link there to this thread so the OP can see my work.
 
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MarkFL

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Feb 24, 2012
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Hello Pillar of Autumn,

we are given to evaluate:

\(\displaystyle I=\int\frac{\sin(\theta)-1}{\cos^2(\theta)}\,d\theta\)

I think the most straightforward method I can think of is to rewrite this as:

\(\displaystyle I=\int \sec(\theta)\tan(\theta)-\sec^2(\theta)\,d\theta\)

Now, given that:

\(\displaystyle \frac{d}{d\theta}\left(\sec(\theta) \right)=\sec(\theta)\tan(\theta)\)

\(\displaystyle \frac{d}{d\theta}\left(\tan(\theta) \right)=\sec^2(\theta)\)

we obtain:

\(\displaystyle I=\sec(\theta)-\tan(\theta)+C\)
 

Prove It

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MHB Math Helper
Jan 26, 2012
1,404
Hello Pillar of Autumn,

we are given to evaluate:

\(\displaystyle I=\int\frac{\sin(\theta)-1}{\cos^2(\theta)}\,d\theta\)

I think the most straightforward method I can think of is to rewrite this as:

\(\displaystyle I=\int \sec(\theta)\tan(\theta)-\sec^2(\theta)\,d\theta\)

Now, given that:

\(\displaystyle \frac{d}{d\theta}\left(\sec(\theta) \right)=\sec(\theta)\tan(\theta)\)

\(\displaystyle \frac{d}{d\theta}\left(\tan(\theta) \right)=\sec^2(\theta)\)

we obtain:

\(\displaystyle I=\sec(\theta)-\tan(\theta)+C\)
Or if you're like me, and never remember the derivative of [tex]\displaystyle \begin{align*} \sec{(\theta)} \end{align*}[/tex], it can be written as [tex]\displaystyle \begin{align*} -\int{ \frac{-\sin{(\theta)}}{\cos^2{(\theta)}} \,d\theta} - \int{\sec^2{(\theta)} \,d\theta} \end{align*}[/tex] and the first integral can be solved with the substitution [tex]\displaystyle \begin{align*} u = \cos{(\theta)} \implies du = -\sin{(\theta)} \, d\theta \end{align*}[/tex].