New to antiderivatives

[scrum]

I just finished my first semester of calc (Got a B)

We covered derivatives and I know that the next 10 weeks of calc are about integrals and antiderivatives, so I figured I'd try to brush up on my calc over christmas and hopefully teach myself the next part of calc to free up my time for Physics when next semester rolls around.

So i've been doing antiderivatives and i get them for the most part but one that i don't understand is.
(x is a theta in the book)

Sin(x) / Cos(x)^2

I got the answer to be 1 / Cos(x) + C (which is correct according to me checking it with a computer) but the way I did it was figuring that the bottom must be Cos(x) since in the quoteint rule of a derivative the bottom is squared, so i imagined differentiating a function with the quoteint rule and got it.

But what would i have done if the term on the bottom was more nasty, say Tan(x)^3 or something where the method I used wouldn't have worked. How would i approach a similar problem.

As always your help is greatly appreciated.

 

[stapel]

But what would i have done if the term on the bottom was more nasty...?

That's what you'll spend a semester learning in Calc 2! :wink:

There are loads of different techniques and formulas to learn, and you'll find out that many (most?) integrals can't even be done! :shock:

Have fun!

Eliz.

 

[Deleted member 4993]

You can also do this by method of substitution.

let

u = cos (x)

\(\displaystyle \frac{du}{dx} = - sin(x)\)

then

\(\displaystyle \int{\frac{sin(x)}{cos^2(x)} dx\)

\(\displaystyle = \int{\frac{-\frac{du}{dx}}{u^2} dx\)

\(\displaystyle =- \int{\frac{du}{u^2}\)

\(\displaystyle = \frac{1}{u} + C\)

\(\displaystyle = \frac{1}{cos(x)} + C\)

You should always check your answer by differentiating.

 

[galactus]

Here's another way using what they call 'integration by parts'. SK's sub method is much easier, but since you want to get ahead of the game...here goes. I hope it's of interest to you. Integration by parts is a biggy. You will definitely see it in Calc II.

\(\displaystyle \int\frac{sin(x)}{cos^{2}(x)}dx=\int{sec^{2}(x)sin(x)}dx\)

Now, let \(\displaystyle u=sin(x), \;\ dv=sec(x)dx, \;\ du=cos(x), \;\ v=tan(x)\)

Using \(\displaystyle uv-\int{vdu}\), put it together:

\(\displaystyle sin(x)tan(x)-\int{tan(x)cos(x)}dx\)

But tan(x)cos(x)=sin(x)

\(\displaystyle sin(x)tan(x)-\int{sin(x)}dx\)

Which gives us:

\(\displaystyle sin(x)tan(x)+cos(x)\)

Now, this boils down to:

\(\displaystyle sin(x)\cdot\frac{sin(x)}{cos(x)}+cos(x)\)

\(\displaystyle \frac{sin^{2}(x)}{cos(x)}+cos(x)\)

\(\displaystyle \frac{sin^{2}(x)+cos^{2}(x)}{cos(x)}\)

As you should know, \(\displaystyle sin^{2}(x)+cos^{2}(x)=1\)

So, we have \(\displaystyle \frac{1}{cos(x)}+C\)

And \(\displaystyle \boxed{\frac{1}{cos(x)}=sec(x)+C}\)

\(\displaystyle \frac{sin(x)}{cos^{2}(x)}\) is also equal to \(\displaystyle sec(x)tan(x)\). You could also go that route.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

If you had something 'nastier' like \(\displaystyle \int\frac{sin(x)}{tan^{3}(x)}dx\)

It isn't as bad as it looks. Use u-substitution.

Rewrite: \(\displaystyle \int\frac{sin(x)}{tan^{3}(x)}dx=\int\frac{cos^{3}(x)}{sin^{2}(x)}dx=\int\frac{(1-sin^{2}(x))cos(x)}{sin^{2}(x)}dx\)

Now, let \(\displaystyle u=sin(x), \;\ du=cos(x)dx\)

Make the subs:

\(\displaystyle \int\frac{1-u^{2}}{u^{2}}du=\int\frac{1}{u^{2}}du-\int{du}\)

Integrating gives: \(\displaystyle u-\frac{1}{u}\)

Resub: \(\displaystyle sin(x)-\frac{1}{sin(x)}=\boxed{sin(x)-csc(x)}\)

You will cover an entire section on these types of problems