Find the General Antiderivative

[alimarie333]

Will someone please help me on these Calculus problems? I have no idea how to solve them!

- Find the general antiderivative (indefinite integral!) for each of the following functions:

I just need help under letter "D", numbers vi, ix and xii. If you have time I would appreciate help on "B" too. Here is a scan of the homework:

http://i16.tinypic.com/4g7xyf4.jpg

Thanks so much for your help. I really appreciate it!
-Alison

 

[galactus]

\(\displaystyle \L\\\int\frac{sin(2x)}{\sqrt{1-sin^{4}(x)}}dx\)

The one is actually easier than it appears.

Let \(\displaystyle \L\\u=sin^{2}(x)\) and \(\displaystyle du=sin(2x)dx\)

Then you have:

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

You should recognize this as arcsin.

\(\displaystyle \L\\sin^{-1}(sin^{2}(x))\)

 

[skeeter]

uhh ... the sheet says you're not to copy anyone else's solutions.

you also ask for help on the only three integrals assigned for turn-in ... tomorrow ... nothing like waiting til the 11th hour.

some hints ...

D. vi ... antiderivative will involve the arctangent function. complete the square in the denominator to "see" the required pattern.

D. ix ... antiderivative will involve the arcsine function. let u = sin²x

D. xii ... a method of solution doesn't jump right out at me for this one. I'd have to take some time to fiddle with it.

B. sketch a right triangle ... let y = arctan(x) be one of the acute angles.
you should know that y = arctan(x) can also be written as tan(y) = x, or better still tan(y) = x/1. with this info, you can label a couple of sides of the right triangle (remember tangent = opposite/adjacent ?) ... finish up by finding value of sin(2y).

 

[Unco]

Here's a start for xii, Ali.

\(\displaystyle \L \mbox{ \int \frac{1}{\sqrt{e^{nx} - 1}} dx = \int \frac{2}{n} \frac{\frac{1}{2}ne^{nx}}{\sqrt{e^{nx} - 1}} \frac{1}{e^{nx}} dx}\)

Your turn: rewrite \(\displaystyle \mbox{{e^{nx}}\) in the denominator in terms of \(\displaystyle \mbox{\sqrt{e^{nx} - 1}}\), so hopefully the appropriate substitution becomes more obvious for you.

 

[soroban]

Hello, alimarie333!

skeeter explained B very nicely . . .

B. For \(\displaystyle x\,>\,0\), find an an algebraic expression for: \(\displaystyle \,\sin(2\,\arctan x)\)

Recall that \(\displaystyle \arctan x\) is some angle \(\displaystyle \theta\).
. . That is: \(\displaystyle \:\theta \:=\:\arctan x\;\;\Rightarrow\;\;\tan\theta \:=\:x\)

Since \(\displaystyle \tan\theta\:=\:\frac{opp}{adj}\), we have this right triangle:

                      *
                  *   |
              *       | x
          *  θ        |
      * - - - - - - - *
              1

Using Pythagorus, we have: \(\displaystyle \:hyp\,=\,\sqrt{x^2\,+\,1}\)

Hence: \(\displaystyle \:\sin\theta \,=\,\frac{x}{\sqrt{x^2\,+\,1}}\;\;\;\cos\theta\,=\,\frac{1}{\sqrt{x^2\,+\,1}}\)

Therefore: \(\displaystyle \L\:\sin(2\theta) \:=\:2\,\sin\theta\cos\theta \:=\:2\left(\frac{x}{\sqrt{x^2\,+\,1}}\right)\left(\frac{1}{\sqrt{x^2\,+\,1}}\right) \:=\:\frac{2x}{x^2\,+\,1}\)