Solving Integral Problem: dx/cos(x) with exp(ix)=t method

[eljose]

I have some problems in calculating the integral dx/cos(x) i have made the change exp(ix)=t and get the solution (2/i)artan(exp(ix)) but what is the real number solution?.thanks

 

[arildno]

It's better to use the half-angle formulas:
$$\cos^{2}(\frac{x}{2})-\sin^{2}(\frac{x}{2})=\cos(x)$$
$$\cos^{2}(\frac{x}{2})+\sin^{2}(\frac{x}{2})=1$$
Your integral is then easy to evaluate with respect to the variable:
$$u=tan(\frac{x}{2})$$

 

[SomeRandomGuy]

I have some problems in calculating the integral dx/cos(x) i have made the change exp(ix)=t and get the solution (2/i)artan(exp(ix)) but what is the real number solution?.thanks

Maybe I am missing something, but isn't dx/cos(x) = sec(x)dx. Doing that integral is very easy, you should be able to handle it.

 

[sinkdeep]

1/cos(x)dx
=cos(x)/(1-sin^2(x))dx
=1/(1-sin^2(x))d(sin(x))

 

[PBRMEASAP]

To SomeRandomGuy,

Do you tutor? I have a kid who could use some help with his math homework. No doubt he could benefit from your penetrating insight and clarity of exposition.

 

[devious_]

To SomeRandomGuy,

Do you tutor? I have a kid who could use some help with his math homework. No doubt he could benefit from your penetrating insight and clarity of exposition.

Here's yet another method to do this integral:

$$\int \sec x \; dx = \int \sec x \; \frac{\sec x+\tan x}{\sec x+\tan x} \; dx$$

$$\int \frac{\sec^{2}x + \sec x\tan x}{\sec x+\tan x}\;dx$$

Take $u=\sec x+\tan x$, then $\frac{du}{dx}=\sec x\tan x+\sec^{2}x$.

$$\int \frac{\sec^{2}x + \sec x\tan x}{\sec x+\tan x} \; dx = \int \frac{1}{u} \; du = \ln(u) \; + \; C = \ln(\sec x+\tan x) \; + \; C$$

 

[SomeRandomGuy]

To SomeRandomGuy,

Do you tutor? I have a kid who could use some help with his math homework. No doubt he could benefit from your penetrating insight and clarity of exposition.

I am one to appreciate sarcasm as I am a very sarcastic individual myself. I am a tutor, actually, and I will rarely complete an integral or a problem for someone I tutor. Instead, I try to have them figure it out for themselves. That way, it is more likely they will remember the process and what to look for.

 

[dextercioby]

Here's yet another method to do this integral:

$$\int \sec x \; dx = \int \sec x \; \frac{\sec x+\tan x}{\sec x+\tan x} \; dx$$

$$\int \frac{\sec^{2}x + \sec x\tan x}{\sec x+\tan x}\;dx$$

Take $u=\sec x+\tan x$, then $\frac{du}{dx}=\sec x\tan x+\sec^{2}x$.

$$\int \frac{\sec^{2}x + \sec x\tan x}{\sec x+\tan x} \; dx = \int \frac{1}{u} \; du = \ln(u) \; + \; C = \ln(\sec x+\tan x) \; + \; C$$

Why complicate ?
$$\int \frac{dx}{\cos x}=\int \frac{d(\sin x)}{1-\sin^2 x}=\frac{1}{2}[\int \frac{d(\sin x)}<br /> {1-\sin x} +\int \frac{d(\sin x)}{1+\sin x}] =...=\frac{1}{2} \ln(\frac{1+\sin x}{1-\sin x})+C<br /> =...=\ln [\frac{\tan(\frac{x}{2})+1}{\tan(\frac{x}{2})-1}]+C.$$

 

[devious_]

dextercioby, your method was already posted. I was only suggesting another method, which I personally don't think is complicated.

 

[dextercioby]

dextercioby, your method was already posted. I was only suggesting another method, which I personally don't think is complicated.

Maybe it was posted...
Check out that horrible trick u pulled in order to integrate secant of x.You wouldn't expect someone to think of it as simple...Obvious...Besides,it requires a lotta derivatives and it uses the secant function which is (as stated by me) unnecessary...
It don't matter really how u pull an integral off.The trick is to do it as elegantly as possible.As comprehendable as possible.I assume u're not a teacher/never be one.

 

[devious_]

Erm.. I wasn't trying to do the integral "as elegantly as possible". I was merely providing another another way to do it, in addition to the two methods already posted...

 

[dextercioby]

Erm.. I wasn't trying to do the integral "as elegantly as possible". I was merely providing another another way to do it, in addition to the two methods already posted...

Well,Mr.Devious,apparently u challanged me:
$$\begin {array} {cc} \int \frac{dx}{\cos x} = \int \sec x dx = \int \sec x (\frac {\tan x}{\tan x })dx<br /> =^{d(\sec x)=\sec x\tan x dx} \int \frac{d(\sec x)}{\sin x \sec x} <br /> \\=^{\sin x=\sqrt{1-\frac{1}{\sec^{2} x}}} \int \frac{d(\sec x)}{\sqrt{\sec^{2} x-1}}<br /> =^{\sec x= \cosh \lambda ; \lambda=\cosh^{-1}(\sec x)}<br /> \int \frac{(\sinh \lambda) d\lambda}{\sinh \lambda}<br /> \\=\lambda +C= \cosh^{-1}(\sec x) +C = \ln(\sec x+\tan x)+C \end{array}$$
,where,of course,the expression $$f^{-1}(x)$$ strands for the inverse of the function $$f(x)$$.It stood for "argcosh (sec x)",but i couldn't find the function,so that's why i used this notation.

Daniel.

 

[devious_]

Challenged you? Why are you taking this personally?

$$\int \sec x \; dx = \int \frac{\cos x}{\cos^{2} x} \; dx = \int \frac{\cos x}{1-\sin^{2} x} \; dx$$

Take $u=\sin x$, then $du=\cos x\;dx$. So:

$$\int \sec x \; dx = \int \frac{1}{1-u^{2}} \; du = \tanh^{-1} u \; + \; C = \tanh^{-1}(\sin x) \; + \; C$$