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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:

$$\int sintsec^2(cost)dt$$

heres what I have so far

$$u=cost$$

$$du=-sintdt$$

$$-\int sec^2(u)du$$

$$-2tan(u) + C$$

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

$$\int \sec^2 x \; dx = \tan x +c$$

~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?

$$\int_{1/2}^{1/6}csc \pi t cot \pi t dt$$

 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? $$\int_{1/2}^{1/6}csc \pi t cot \pi t dt$$
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..
$$\int_{1/2}^{1/6} \frac{1}{sin\pi t) \frac{cos \pi t}{sin \pi t}dt$$
ended up with...
$$\int_{1/2}^{1/6} cot \pi t$$

im guessing its wrong haha

 Hootenanny May26-06 03:07 PM

Quote:
 Quote by suspenc3 ok..so i did.. $$\int_{1/2}^{1/6} \frac{1}{sin\pi t) \frac{cos \pi t}{sin \pi t}dt$$ ended up with... $$\int_{1/2}^{1/6} cot \pi t$$ im guessing its wrong haha
It's almost there :wink:

$$\frac{1}{\sin \pi t} \cdot \frac{\cos \pi t}{\sin \pi t} = \frac{\cos \pi t}{\sin^2 \pi t}$$

Now, see what you can do with the identity $\sin^2 \theta + \cos^2 \theta = 1$. 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: $$\int sintsec^2(cost)dt$$ heres what I have so far $$u=cost$$ $$du=-sintdt$$ $$-\int sec^2(u)du$$ $$-2tan(u) + C$$ 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:

$$\int sintsec^2(cost)dt= 2tan(cos(t)+ C$$

 EbolaPox May26-06 06:25 PM

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

$$\int sin(t) sec (cos(x))^2 dx$$

$$u = cos(x) du = -sin(x) dx$$ This is right. Substituting back yields

$$- \int sec(u)^2 du$$ 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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