Trig Integration Problem

[Jason76]

\(\displaystyle \int \sec^{2} x \tan x dx\)

\(\displaystyle \int \sec^{2} x (u) dx\)

\(\displaystyle u = \tan x\)

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

\(\displaystyle \int u(du)\) - Confused on this. Making the right moves?

 

[Jason76]

\(\displaystyle \int \sec^{2} x \tan x dx\)

\(\displaystyle \int \sec^{2} x (u) dx\)

\(\displaystyle u = \tan x\)

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

\(\displaystyle \int u(du)\)

\(\displaystyle \int (\tan x)(du)\)

\(\displaystyle \rightarrow \ln(\sec) + C\)

 

[daon2]

Ok I see now. tan²(x) = sec²(x? - 1 so the the two are identical except for a constant that gets rolled into C.

This is the first "go home and think about it" problem I give my students when doing substitution.

 

[JeffM]

\(\displaystyle \int \sec^{2} x \tan x dx\)

\(\displaystyle \int \sec^{2} x (u) dx\)

\(\displaystyle u = \tan x\)

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

\(\displaystyle \int u(du)\)

\(\displaystyle \int (\tan x)(du)\) You can't integrate tan(x) against du. Variable needs to be consistent.

\(\displaystyle \rightarrow \ln(\sec) + C\)

\(\displaystyle u = tan(x) \implies \dfrac{du}{dx} = sec^2(x) \implies du = sec^2(x)\ dx \implies \displaystyle \int sec^2(x) * tan(x)\ dx = \int u\ du =\)

\(\displaystyle \dfrac{u^2}{2} + C = \dfrac{tan^2(x)}{2} + C = \dfrac{tan^2(x) + 1}{2} + C - \dfrac{1}{2} = \dfrac{sec^2(x)}{2} + K.\)

 

[Jason76]

All this identity stuff isn't needed, as this is a straight integral u substitution problem. maybe

\(\displaystyle \int \sec^{2} x \tan x dx\)

\(\displaystyle u = \sec x\)

\(\displaystyle du = \sec x \tan x dx\)

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

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

Using the formula \(\displaystyle \dfrac{u^{n + 1}}{n + 1}\) which we would only use in the case of trig function raised to some power.

\(\displaystyle \rightarrow (\dfrac{1}{\sec x})\dfrac{u}{3} + C\)

\(\displaystyle \rightarrow (\dfrac{1}{\sec x})\dfrac{\sec^{3} x}{3} + C\)

 

[Jason76]

OR we could do it this way (This flows much better):

\(\displaystyle \int \sec^{2} x \tan x dx\)

\(\displaystyle u = \tan x\)

\(\displaystyle du = \sec^{2} x dx\) - exact match. No need for manipulation.

\(\displaystyle \int u^{2} du\)

\(\displaystyle \rightarrow \dfrac{u^{3}}{3} + C\)

\(\displaystyle \rightarrow \dfrac{\tan ^{3} x}{3} + C\) Maybe this is right.

HOWEVER, if we made \(\displaystyle u = \sec x\)

\(\displaystyle du = \sec x \tan x \) etc.. and went that route, then, shouldn't it come up with the same answer?

 

[Jason76]

Professor said to use u substitution:

\(\displaystyle u = \tan x\)

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

 

[Jason76]

\(\displaystyle \int \sec^{2} x \tan x dx\)

\(\displaystyle u = \tan x\)

\(\displaystyle du = \sec^{2} x dx\) - exact match. No need for manipulation.

\(\displaystyle \int u^{2} du\)

\(\displaystyle \rightarrow \dfrac{u^{3}}{3} + C\)

\(\displaystyle \rightarrow \dfrac{\tan ^{3} x}{3} + C\)

 

[JeffM]

OR we could do it this way (This flows much better):

\(\displaystyle \int \sec^{2} x \tan x dx\)

\(\displaystyle u = \tan x\)

\(\displaystyle du = \sec^{2} x dx\) - exact match. No need for manipulation.

\(\displaystyle \int u^{2} du\)

\(\displaystyle \rightarrow \dfrac{u^{3}}{3} + C\)

\(\displaystyle \rightarrow \dfrac{\tan ^{3} x}{3} + C\) Maybe this is right.

HOWEVER, if we made \(\displaystyle u = \sec x\)

\(\displaystyle du = \sec x \tan x \) etc.. and went that route, then, shouldn't it come up with the same answer?

As has now been explained several times, either way works if you do it correctly. But you have not done it correctly above.

\(\displaystyle u = tan(x) \implies du = sec^2(x)dx \implies \displaystyle \int sec^2(x) * tan(x)\ dx = \int tan(x) * (sec^2(x)\ dx) = \int u\ du,\ not \int u^2\ du.\)

So the answer is \(\displaystyle \dfrac{u^2}{2} + C_1 = \dfrac{tan^2(x)}{2} + C_1.\)

Do it the other way

\(\displaystyle u = sec(x) \implies du = sec(x) * tan(x)dx \implies \displaystyle \int sec^2(x) * tan(x)\ dx = \int sec(x) * (sec(x) * tan(x)\ dx) = \int u\ du = \)

\(\displaystyle \dfrac{u^2}{2} + C_2 = \dfrac{sec^2(x)}{2} + C_2.\)

But those two answers mean the same thing.

\(\displaystyle \dfrac{sec^2(x)}{2} + C_2= \dfrac{tan^2(x) + 1}{2} = \dfrac{tan^2(x)}{2} + \dfrac{1}{2} + C_2 = \dfrac{tan^2(x)}{2} + C_1.\)

 

[Deleted member 4993]

OR we could do it this way (This flows much better):

\(\displaystyle \int \sec^{2} x \tan x dx\)

\(\displaystyle u = \tan x\)

\(\displaystyle du = \sec^{2} x dx\) - exact match. No need for manipulation.

\(\displaystyle \int u^{2} du\)......................... Incorrect - how did you come to this line from above?!!

\(\displaystyle \rightarrow \dfrac{u^{3}}{3} + C\)

\(\displaystyle \rightarrow \dfrac{\tan ^{3} x}{3} + C\) Maybe this is right.

HOWEVER, if we made \(\displaystyle u = \sec x\)

\(\displaystyle du = \sec x \tan x \) etc.. and went that route, then, shouldn't it come up with the same answer?

.

 

[Jason76]

\(\displaystyle \int \sec^{2} x \tan x dx\)

\(\displaystyle u = \tan x\)

\(\displaystyle du = \sec^{2} x dx\) - exact match. No need for manipulation.

\(\displaystyle \int u du\)

\(\displaystyle \rightarrow \ln \sec x + C\)

 

[Deleted member 4993]

\(\displaystyle \int \sec^{2} x \tan x dx\)

\(\displaystyle u = \tan x\)

\(\displaystyle du = \sec^{2} x dx\) - exact match. No need for manipulation.

\(\displaystyle \int u du\)

\(\displaystyle \rightarrow \ln \sec x + C\) ............... Incorrect ...... how did you get that from line above??!!

\(\displaystyle \int u du\) = \(\displaystyle \displaystyle{\frac{1}{2}u^2 + C \ = \ \frac{1}{2}tan^2(x) + C}\)

.

 

[JeffM]

\(\displaystyle \int \sec^{2} x \tan x dx\)

\(\displaystyle u = \tan x\)

\(\displaystyle du = \sec^{2} x dx\) - exact match. No need for manipulation.

\(\displaystyle \int u du\)

\(\displaystyle \rightarrow \ln \sec x + C\)

Jason

The whole point of u-substitution is to reduce a complicated integral to a simple one.

You integrate the simple one USING the substitution and get an ANSWER in u.

Then you substitute back from u to x.

\(\displaystyle \displaystyle \int u\ du = \dfrac{u^2}{2} + C.\) I KNOW you know that.

Integration is done now.

Just substitute back. In this case u = tan(x) so the answer is \(\displaystyle \dfrac{tan^2(x)}{2} + C.\)

You are making what is very simple hard by not first finding the integral in terms of u and then converting to x in a separate step.

What can be hard about u-substitution is finding the correct substitution in the first place. You want to find u = g(x) such that

\(\displaystyle f(x)\ dx = h(u)\ du,\) where it is easy to integrate h(u) du.

 

[Jason76]

Jason

The whole point of u-substitution is to reduce a complicated integral to a simple one.

You integrate the simple one USING the substitution and get an ANSWER in u.

Then you substitute back from u to x.

\(\displaystyle \displaystyle \int u\ du = \dfrac{u^2}{2} + C.\) I KNOW you know that.

Integration is done now.

Just substitute back. In this case u = tan(x) so the answer is \(\displaystyle \dfrac{tan^2(x)}{2} + C.\)

You are making what is very simple hard by not first finding the integral in terms of u and then converting to x in a separate step.

What can be hard about u-substitution is finding the correct substitution in the first place. You want to find u = g(x) such that

\(\displaystyle f(x)\ dx = h(u)\ du,\) where it is easy to integrate h(u) du.

I see what's going on, but isn't the integral of \(\displaystyle \tan = \ln|sec|\)?? How does that play in this?

 

[Deleted member 4993]

I see what's going on, but isn't the integral of \(\displaystyle \tan = \ln|sec|\)?? How does that play in this?

It does not -

you are not integrating tan(x) dx here

You are integrating tan(x) * sec²(x) dx