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

[MarkFL]

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.

 

[MarkFL]

Hello Pillar of Autumn,

we are given to evaluate:

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

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

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

Now, given that:

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

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

we obtain:

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

 

[Prove It]

Hello Pillar of Autumn,

we are given to evaluate:

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

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

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

Now, given that:

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

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

we obtain:

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

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