1. ## Trig Integral

$\int SEC^4{5x} = \int ({SEC^2{5x}})({SEC^2{5x}})$
Then
$\int (1+TAN^2{5x})(SEC^2{5x})$
Then I multiply out
$\int (SEC^2{5x})+(SEC^2{5x}TAN^2{5x})$

I would assume U sub at this point
$U= SEC dU= SECTAN$
But I am not sure, what or how I should bring along with the SEC

Any help?

thanks
Rich

2. Originally Posted by Spoolx
$\int SEC^4{5x} = \int ({SEC^2{5x}})({SEC^2{5x}})$
Then
$\int (1+TAN^2{5x})(SEC^2{5x})$
Then I multiply out
$\int (SEC^2{5x})(SEC^2{5x}TAN^2{5x})$
Fine so far, except you're missing a plus sign in the last line, which I assume is a typo. Now to substitute, which is really in order to work backwards through the chain rule.

Just in case a picture helps...

... where (key in first spoiler) ...

Spoiler:

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case ), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

The general drift is...

Spoiler:
$\int 1 + u^2\ du = u + \frac{1}{3} u^3$

and so we can explicate F...

Hope this helps - if not, someone will probably show the sub with differentials. However, I say...
_________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!

3. I edited the original to include the + sign, thanks for catching my mistake

4. $\displaystyle\int{sec^4(5x)}dx=\int{sec^2(5x)sec^2 (5x)}dx=\int{\left[1+tan^2(5x)\right]sec^2(5x)}dx$

Substitute

$u=tan(5x)$

$\displaystyle\Rightarrow\frac{du}{dx}=5sec^2(5x)$

by the chain rule

$\displaystyle\Rightarrow\ sec^2(5x)dx=\frac{du}{5}$

The integral becomes

$\displaystyle\frac{1}{5}\int{\left[1+u^2\right]du$

5. Hello, Rich!

A rehash of Archie's solution . . .

$\displaystyle \int \sec^4\!5x\,dx \:= \:\int (\sec^2\!5x)(\sec^2\!5x)\,dx$

Then: . $\displaystyle \int (1+\tan^2\!5x)(\sec^2\!5x)\,dx$

Then I multiply out: . $\displaystyle \int (\sec^2\!5x +\sec^2\!5x\tan^2\!5x)\,dx$

$\displaystyle \text{You have: }\;\int\sec^2\!5x\,dx + \int(\tan 5x)^2(\sec^2\!5x\,dx)$

For the first integral, use: . $\displaystyle \int \sec\theta\,d\theta \:=\:\tan\theta + C$

For the second integral, let: . $u \,=\, \tan x \quad\Rightarrow\quad du \,=\,\sec^2\!x\,dx$

. . and you have: . $\displaystyle \int u^2\,du \:=\:\tfrac{1}{3}u^3 + C$

Got it?

6. Originally Posted by Soroban
Hello, Rich!

A rehash of Archie's solution . . .

$\displaystyle \text{You have: }\;\int\sec^2\!5x\,dx + \int(\tan 5x)^2(\sec^2\!5x\,dx)$

For the first integral, use: . $\displaystyle \int \sec\theta\,d\theta \:=\:\tan\theta + C$

For the second integral, let: . $u \,=\, \tan x \quad\Rightarrow\quad du \,=\,\sec^2\!x\,dx$

. . and you have: . $\displaystyle \int u^2\,du \:=\:\tfrac{1}{3}u^3 + C$

Got it?
So to refresh my algebra

$tan^2{5x} = (tan5x)^2$ ?

Also above, is your U supposed to be?
$u= tan5x$ ?
if not where does the 5 go?

7. Originally Posted by Spoolx
$tan^2{5x} = (tan5x)^2$ ? Mr F says: Yes.
$u= tan5x$ ? Mr F says: Yes. Soroban made a simple typo. You ought to be able to make the appropriate corrections in the calculations.