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-   -   Substitution (http://www.physicsforums.com/showthread.php?t=122079)

suspenc3 May26-06 02:27 PM

Substitution
 
Hi, Im having a little bit of trouble with the following:

[tex]\int sintsec^2(cost)dt[/tex]

heres what I have so far

[tex]u=cost[/tex]

[tex]du=-sintdt[/tex]

[tex]-\int sec^2(u)du[/tex]

[tex]-2tan(u) + C[/tex]

is this right?

vsage May26-06 02:30 PM

As I recall the derivative of tan(u) is sec^2(u)du

suspenc3 May26-06 02:33 PM

yea but im integrating..so the antiderivative of sec^2(u) is tanu?

Hootenanny May26-06 02:36 PM

[tex]\int \sec^2 x \; dx = \tan x +c[/tex]

~H

suspenc3 May26-06 02:42 PM

I have another..i dont know where to start..can someone point out what I should sub U for?

[tex]\int_{1/2}^{1/6}csc \pi t cot \pi t dt[/tex]

Hootenanny May26-06 02:51 PM

Quote:

Quote by suspenc3
I have another..i dont know where to start..can someone point out what I should sub U for?

[tex]\int_{1/2}^{1/6}csc \pi t cot \pi t dt[/tex]

My first step would be to turn the cosec and cot into sine and cosine. See where that takes you.

~H

suspenc3 May26-06 03:01 PM

ok..so i did..
[tex] \int_{1/2}^{1/6} \frac{1}{sin\pi t) \frac{cos \pi t}{sin \pi t}dt[/tex]
ended up with...
[tex] \int_{1/2}^{1/6} cot \pi t[/tex]

im guessing its wrong haha

Hootenanny May26-06 03:07 PM

Quote:

Quote by suspenc3
ok..so i did..
[tex] \int_{1/2}^{1/6} \frac{1}{sin\pi t) \frac{cos \pi t}{sin \pi t}dt[/tex]
ended up with...
[tex] \int_{1/2}^{1/6} cot \pi t[/tex]

im guessing its wrong haha

It's almost there :wink:

[tex]\frac{1}{\sin \pi t} \cdot \frac{\cos \pi t}{\sin \pi t} = \frac{\cos \pi t}{\sin^2 \pi t}[/tex]

Now, see what you can do with the identity [itex]\sin^2 \theta + \cos^2 \theta = 1[/itex]. Btw, I think this can be done without a substitution.

~H

suspenc3 May26-06 03:14 PM

do you mean..sin^2(pi t) = 1-cos^2(pi t)..and then sub?

arunbg May26-06 03:25 PM

Forget about the subs.
Try putting sin(pi*t) = x
What is dx ?

HallsofIvy May26-06 05:58 PM

Quote:

Quote by suspenc3
Hi, Im having a little bit of trouble with the following:

[tex]\int sintsec^2(cost)dt[/tex]

heres what I have so far

[tex]u=cost[/tex]

[tex]du=-sintdt[/tex]

[tex]-\int sec^2(u)du[/tex]

[tex]-2tan(u) + C[/tex]

is this right?

As others have pointed out, the anti-derivative of sec2(u) is tan(u), not -tan(u).

Also, the original problem does not say anything about "u"! That was your "invention". To properly answer the question, you need to go back to t:

[tex]\int sintsec^2(cost)dt= 2tan(cos(t)+ C[/tex]

EbolaPox May26-06 06:25 PM

Actually, the original poster (with respect to the original question) is correct, save for the two.

[tex] \int sin(t) sec (cos(x))^2 dx [/tex]

[tex] u = cos(x) du = -sin(x) dx [/tex] This is right. Substituting back yields

[tex] - \int sec(u)^2 du [/tex] Just as the OP said. The negative sign is the result of the du = -sin(x) dx part.

Now, the antiderivative of sec(x)^2 = tan(x) + C (as everyone as stated)

- ( tan(u) + C)
-tan(cos(x) + C.

The original poster's only error is the two infront. The minus sign is correct . Check with differentiatoin (or use the "Integrator").


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