Trig Integral

[Jason76]

Trig derivatives a piece of cake, but these are different:

\(\displaystyle \int \tan(2x) dx\)

\(\displaystyle u = \sec^{2}(2x)\)

\(\displaystyle du = 2 \sec^{2}(2x) dx\)

\(\displaystyle \dfrac{1}{2} du = \sec^{2}(2x) dx\)

\(\displaystyle \dfrac{1}{2} \int ? \)

The end answer (after all work) should be something similar to \(\displaystyle \ln|\sec(x)| + C\) since that's the integral of \(\displaystyle \tan(x)\)

Of course, i might doing some big goof here. Feel free to point out.

 

[pappus]

Trig derivatives a piece of cake, but these are different:

\(\displaystyle \int \tan(2x) dx\)

\(\displaystyle u = \sec^{2}(2x)\)

\(\displaystyle du = 2 \sec^{2}(2x)\)

What's the next step? hint

The end answer (after all work) should be something similar to \(\displaystyle \ln|\sec(x)| + C\) since that's the integral of \(\displaystyle \tan(x)\)

Of course, i might doing some big goof here. Feel free to point out.

Hello,

that's the way I would do this problem:

\(\displaystyle \int \left(\tan(2x)\right) dx = \int \left(\frac{\sin(2x)}{\cos(2x)}\right) dx\)

Then use the substitution:

\(\displaystyle u = \cos(2x)~\implies~du = -2 \cdot \sin(2x) dx\)

That means:

\(\displaystyle \int \left( \tan(2x) \right) dx = -\frac12 \int \left( \frac{-2 \cdot \sin(2x)}{\cos(2x)} \right) dx \)

Can you finish it from here?

 

[Jason76]

Hello,

that's the way I would do this problem:

\(\displaystyle \int \left(\tan(2x)\right) dx = \int \left(\frac{\sin(2x)}{\cos(2x)}\right) dx\)

Then use the substitution:

\(\displaystyle u = \cos(2x)~\implies~du = -2 \cdot \sin(2x) dx\)

That means:

\(\displaystyle \int \left( \tan(2x) \right) dx = -\frac12 \int \left( \frac{-2 \cdot \sin(2x)}{\cos(2x)} \right) dx \)

Can you finish it from here?

Looks interesting, but \(\displaystyle \tan(x)\) has an integral (without converting to \(\displaystyle \sin(x)\) and \(\displaystyle \cos(x)\) But knowing both ways would be great.

 

[Deleted member 4993]

Looks interesting, but \(\displaystyle \tan(x)\) has an integral (without converting to \(\displaystyle \sin(x)\) and \(\displaystyle \cos(x)\) But knowing both ways would be great.

\(\displaystyle \displaystyle \frac{d}{dx}\left[ln[sec(x)]\right ] \ = \ tan(x)\)

 

[Deleted member 4993]

Trig derivatives a piece of cake, but these are different:

\(\displaystyle \int \tan(2x) dx\)

\(\displaystyle u = \sec^{2}(2x)\) ← Why are you making that substitution?

\(\displaystyle du = 2 \sec^{2}(2x) dx\)

\(\displaystyle \dfrac{1}{2} du = \sec^{2}(2x) dx\)

\(\displaystyle \dfrac{1}{2} \int ? \)

The end answer (after all work) should be something similar to \(\displaystyle \ln|\sec(x)| + C\) since that's the integral of \(\displaystyle \tan(x)\)

Of course, i might doing some big goof here. Feel free to point out.

It is not clear what you are trying to do!!

 

[Jason76]

1st way:

\(\displaystyle \int \tan(2x) dx\)

\(\displaystyle u = \tan(2x)\)

\(\displaystyle du = 2 * \sec^{2}(2x)\)

\(\displaystyle \dfrac{1}{2}du = \sec^{2}(2x)\)

\(\displaystyle \dfrac{1}{2}\int du\)

\(\displaystyle \dfrac{1}{2}{\tan 2x} + C\)

I know this isn't right. But this following the pattern of the \(\displaystyle a^{u}\) integral problem.

2nd way:

\(\displaystyle \int \tan(2x) dx\)

\(\displaystyle u = 2x\)

\(\displaystyle du = 2 dx \rightarrow du(\dfrac{1}{2}) = dx \)

\(\displaystyle \dfrac{1}{2}\int \tan (u)(du) \)

\(\displaystyle \dfrac{1}{2} \ln |\sec(2x)| + C\)

But one book was saying the integral of \(\displaystyle \tan(x)\) was \(\displaystyle -\ln cos(x)\) or is it both?

 

[lookagain]

2nd way:

\(\displaystyle \int \tan(2x) dx\)

\(\displaystyle u = 2x\)

\(\displaystyle du = 2 dx \rightarrow du(\dfrac{1}{2}) = dx \)

\(\displaystyle \dfrac{1}{2}\int \tan (u)(du) \)

\(\displaystyle \dfrac{1}{2} \ln |\sec(2x)| + C\) \(\displaystyle \ \ \ \ \) <------ Yes.

But one book was saying the integral of \(\displaystyle \tan(x)\) was \(\displaystyle -\ln cos(x)\) or is it both?

\(\displaystyle \int \tan(x) dx \ = \ -\ln|cos(x)| \ + C \ \ \ \ or \ \ \ \ \ln|sec(x)| \ + \ C.\)

 

[Deleted member 4993]

1st way:

\(\displaystyle \int \tan(2x) dx[/tex ]..............................................given .......................(1)

\(\displaystyle u = \tan(2x)\)

\(\displaystyle du = 2 * \sec^{2}(2x)\)

\(\displaystyle \dfrac{1}{2}du = \sec^{2}(2x)\)

\(\displaystyle \dfrac{1}{2}\int du\) ← This is equivalent to \(\displaystyle \displaystyle \int sec^2(2x) dx \) NOT what you started with in line (1)

\(\displaystyle \dfrac{1}{2}{\tan 2x} + C\)......................... This is NO-WAY

I know this isn't right. But this following the pattern of the \(\displaystyle a^{u}\) integral problem.

2nd way:

\(\displaystyle \int \tan(2x) dx\)

\(\displaystyle u = 2x\)

\(\displaystyle du = 2 dx \rightarrow du(\dfrac{1}{2}) = dx \)

\(\displaystyle \dfrac{1}{2}\int \tan (u)(du) \)

\(\displaystyle \dfrac{1}{2} \ln |\sec(2x)| + C\)

But one book was saying the integral of \(\displaystyle \tan(x)\) was \(\displaystyle -\ln cos(x)\) or is it both?\)

\(\displaystyle .\)

 

[Jason76]

So I'm assuming that the strategy for the "a to the u integral" thread doesn't work with a trig integral. You cannot make \(\displaystyle \tan(2x) = u\). You can only make \(\displaystyle 2x = u\).

http://www.freemathhelp.com/forum/threads/83129-a-to-the-u-Intergal

\(\displaystyle \int_{1}^{5} 3^{2x} dx\).

\(\displaystyle u = 3^{2x}\).

\(\displaystyle du = 3^{2x}(\ln 3)(2) dx\).

\(\displaystyle \dfrac{du}{ 2\ln 3} = 3^{2x} dx\).

\(\displaystyle \dfrac{1}{2 \ln 3} \int_{9}^{59049} u(du)\).

\(\displaystyle \dfrac{u}{2\ln 3} |_{9}^{59049} \).

\(\displaystyle \dfrac{59049}{2\ln 3} - \dfrac{9}{2\ln 3} = \dfrac{59040}{2\ln 3} = \dfrac{29520}{\ln 3}\).

 

[Deleted member 4993]

So I'm assuming that the strategy for the "a to the u integral" thread doesn't work with a trig integral. You cannot make \(\displaystyle \tan(2x) = u\). You can only make \(\displaystyle 2x = u\).

http://www.freemathhelp.com/forum/threads/83129-a-to-the-u-Intergal

\(\displaystyle \int_{1}^{5} 3^{2x} dx\).

\(\displaystyle u = 3^{2x}\).

\(\displaystyle du = 3^{2x}(\ln 3)(2) dx\).

\(\displaystyle \dfrac{du}{ 2\ln 3} = 3^{2x} dx\).

\(\displaystyle \dfrac{1}{2 \ln 3} \int_{9}^{59049} u(du)\).← How many times do we have to point out that this is INCORRECT

\(\displaystyle \dfrac{u}{2\ln 3} |_{9}^{59049} \).

\(\displaystyle \dfrac{59049}{2\ln 3} - \dfrac{9}{2\ln 3} = \dfrac{59040}{2\ln 3} = \dfrac{29520}{\ln 3}\)

.

 

[Jason76]

.

Sorry, forgot to edit my posts in that regard.