Stumped on trig integral

**x0ne** · May 10th 2009, 02:13 PM

Problem:
$\int \frac {1}{\cos u\cot u} du$

So I am kind of stumped with these Us in the equation. Our professor has been gracious enough to point us to an integral solving site, but I get different answers based on switching U for X.

I don't think I need to switch any of the variables out, but I am trying to figure out where I want to go from here. It doesn't seem like I should use the substitution method on this, but I could be wrong. Could someone walk me through this?

**TheEmptySet** · May 10th 2009, 02:18 PM

Originally Posted by x0ne

Problem:
$\int \frac {1}{\cos u\cot u} du$

So I am kind of stumped with these Us in the equation. Our professor has been gracious enough to point us to an integral solving site, but I get different answers based on switching U for X.

I don't think I need to switch any of the variables out, but I am trying to figure out where I want to go from here. It doesn't seem like I should use the substitution method on this, but I could be wrong. Could someone walk me through this?

$\int \frac{1}{\cos(u)\cot(u)}du=\int\frac{1}{\cos(u)\fr ac{\cos(u)}{\sin(u)}}du=\int \frac{\sin(u)}{\cos^2(u)}du$

now this is a substitution let $t=\cos(u) \implies dt=-\sin(u)du$

So we get

$\int \frac{-1}{t^2}dt=\int -t^{-2}dt=t^{-1}+c=\frac{1}{\cos(u)}+c=\sec(u)+c$

**Plato** · May 10th 2009, 02:19 PM

$\frac{1}<br /> {{\cos (u)\cot (u)}} = \frac{{\sin (u)}}<br /> {{\cos ^2 (u)}}$

**x0ne** · May 10th 2009, 02:26 PM

Originally Posted by Plato

$\frac{1}<br /> {{\cos (u)\cot (u)}} = \frac{{\sin (u)}}<br /> {{\cos ^2 (u)}}$

On this site (Wolfram Mathematica Online Integrator) it says the answer should be:
$x \sec u \tan u$

I know from the trig identities that sin/cos = tan and that would then leave 1/cos which is sec. My only gripe with their answer is that X that is upfront. Do you know how that is calculated?

**TheEmptySet** · May 10th 2009, 02:42 PM

Originally Posted by x0ne

On this site (Wolfram Mathematica Online Integrator) it says the answer should be:
$x \sec u \tan u$

I know from the trig identities that sin/cos = tan and that would then leave 1/cos which is sec. My only gripe with their answer is that X that is upfront. Do you know how that is calculated?

No offense but I think you used it incorrectly.

First where did the x come from? There are no x's in the original and you are not integrating with respect to x, but with respect to u.

You can check that the answer I gave you is correct by taking its derviative.

$\frac{d}{du}\sec{u}=sec(u)\tan(u)=\frac{\sin(u)}{\ cos(u)\cos(u)}=\frac{1}{\cos(u)\cot(u)}$

The last step comes from multiplying the top and bottom of the fraction by

$\frac{1}{\sin(u)}$

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