Indefinite Integrals (Antiderivatives)

**Kristenx2** · December 5th 2011, 06:08 AM

Hi, Kristenx2 again. Same take-home test.
I missed the day we did antiderivatives, but is seems easy enough... I hope.
My only issue is when it comes to the chain rule... Trig functions screw me up.
I would just appreciate someone checking my work for the first, and possibly showing me how to do the second. I know I should use the quotient rule, but really.
*Since I can't make that fancy antiderivative squiggle line, I will substitute with / *

(a) /(3secxtanx+7sec^2x)dx
=3/2sec^2x+7tanx+c

(b) /(3x^3+5x^7)
--------------------------- dx
(square root of x)

= ?

**Quacky** · December 5th 2011, 06:36 AM

Originally Posted by Kristenx2

Hi, Kristenx2 again. Same take-home test.
I missed the day we did antiderivatives, but is seems easy enough... I hope.
My only issue is when it comes to the chain rule... Trig functions screw me up.
I would just appreciate someone checking my work for the first, and possibly showing me how to do the second. I know I should use the quotient rule, but really.
*Since I can't make that fancy antiderivative squiggle line, I will substitute with / *

(a) /(3secxtanx+7sec^2x)dx
=3/2sec^2x+7tanx+c

(b) /(3x^3+5x^7)
--------------------------- dx
(square root of x)

= ?

Where does the chain rule come into this?

For a) Differentiate $sec(x)+tan(x)$ . You should notice something which should make the integration easier.

For b), we have:

$\int\frac{3x^3+5x^7}{\sqrt{x}}}dx$

$=\int (\frac{3x^3}{\sqrt{x}}+\frac{5x^7}{\sqrt{x}}})dx$

$=\int (3x^{\frac{5}{2}}+5x^{\frac{13}{2}})~dx$

**Kristenx2** · December 5th 2011, 07:24 AM

Oh! Thank you for your help! As you can see, I'm horribly confused, so I'm very unsure as to where anything comes in.

I was looking on other sites as to how to do antiderivatives when there is a quotient, and they just said to flip it up.
So I have
/(3x^3+5x^7)(x^-1/2)dx.
And ended up with
6/7x^7/2+1/3x^15/2+c. I just used the power rule. I can use that even when there are constants, right?

**Quacky** · December 5th 2011, 07:52 AM

I think you really need a catch up session with your teacher. Also, for part b) I did most of the work for you. It should be $\frac{6}{7}x^{\frac{7}{2}}+\frac{2}{3}x^\frac{15}{ 2}+C$ Did you manage to get anywhere with part a)?

**Kristenx2** · December 5th 2011, 08:00 AM

Me, too. I noticed you got 2/3 where I got 1/3, that's probably just a mistake in my math.
For the first one, would it just be
3sec^2x+7tanx+c?
Just leave the constants and figure out what makes the trig functions?

**Quacky** · December 5th 2011, 08:10 AM

No, did you try what I suggested in post #2?

**Kristenx2** · December 5th 2011, 08:22 AM

Yes, I got sec(x)tan(x)+sec^2(x), which does look an awful lot like the original problem!
However, how do I get the coefficients to come into play?

**Quacky** · December 5th 2011, 08:28 AM

The coefficients don't change, as we have:

$\int(3sec(x)tan(x)+7sec^2(x))~dx$

$\int{3}sec(x)tan(x)~dx+\int{7sec^2(x)}~dx$

$=3\int{sec(x)tan(x)}~dx + 7\int{sec^2(x)}~dx$

**Prove It** · December 5th 2011, 04:32 PM

Originally Posted by Kristenx2

Hi, Kristenx2 again. Same take-home test.
I missed the day we did antiderivatives, but is seems easy enough... I hope.
My only issue is when it comes to the chain rule... Trig functions screw me up.
I would just appreciate someone checking my work for the first, and possibly showing me how to do the second. I know I should use the quotient rule, but really.
*Since I can't make that fancy antiderivative squiggle line, I will substitute with / *

(a) /(3secxtanx+7sec^2x)dx
=3/2sec^2x+7tanx+c

(b) /(3x^3+5x^7)
--------------------------- dx
(square root of x)

= ?

This is just a matter of personal preference, but I don't like having to remember lots of integration formulas. For trigonometric functions I really only ever remember $\displaystyle \begin{align*} \int{\cos{x}\,dx} = \sin{x} + C, \int{\sin{x}\,dx} = -\cos{x} + C \end{align*}$ and $\displaystyle \begin{align*} \int{\sec^2{x}\,dx} = \tan{x} + C \end{align*}$ . Every other time I'll try to convert the integrand into some combination of these...

$\displaystyle \begin{align*} \int{3\sec{x}\tan{x} + 7\sec^2{x}\,dx} &= 3\int{\sec{x}\tan{x}\,dx} + 7\int{\sec^2{x}\,dx} \\ &= 3\int{\frac{1}{\cos{x}} \cdot \frac{\sin{x}}{\cos{x}}\,dx} + 7\tan{x} + C \\ &= 3\int{\cos^{-2}{x}\sin{x}\,dx} + 7\tan{x} + C \\ &= -3\int{\cos^{-2}{x}\left(-\sin{x}\right) dx} + 7\tan{x} + C \end{align*}$

Now make the substitution $\displaystyle \begin{align*} u = \cos{x} \implies du = -\sin{x}\,dx \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} -3\int{\cos^{-2}{x}\left(-\sin{x}\right) dx} + 7\tan{x} + C &= -3\int{u^{-2}\,du} + 7\tan{x} + C \\ &= \frac{-3u^{-1}}{-1} + 7\tan{x} + C \\ &= \frac{3}{\cos{x}} + 7\tan{x} + C \\ &= 3\sec{x} + 7\tan{x} + C\end{align*}$

which is the same answer you would get using the other method as well, seeing as $\displaystyle \begin{align*} \frac{d}{dx}\left(\sec{x}\right) = \sec{x}\tan{x} \end{align*}$ as you found before

**mr fantastic** · December 5th 2011, 05:17 PM

Is a take-home test meant to be your own work?

Thread closed. See rule #6: http://www.mathhelpforum.com/math-he...hp?do=vsarules.

Thread closed.

Originally Posted by Kristenx2

Hi, Kristenx2 again. Same take-home test.
I missed the day we did antiderivatives, but is seems easy enough... I hope.
My only issue is when it comes to the chain rule... Trig functions screw me up.
I would just appreciate someone checking my work for the first, and possibly showing me how to do the second. I know I should use the quotient rule, but really.
*Since I can't make that fancy antiderivative squiggle line, I will substitute with / *

(a) /(3secxtanx+7sec^2x)dx
=3/2sec^2x+7tanx+c

(b) /(3x^3+5x^7)
--------------------------- dx
(square root of x)

= ?

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