MHB Integral- Without using integration technnique

[Chipset3600]

Hello MHB, how can i solve this without use integration technniques...

\int tan(t)sec^3(t)dt

 

[SuperSonic4]

Re: Integral- Without use integration technnique

Hello MHB, how can i solve this without use integration technniques...

\int tan(t)sec^3(t)dt

How do you mean "without using integration techniques"? Surely you need integration techniques to solve an integral? Also what have you tried?

Hint: Let $u = \sec(t) = \frac{1}{\cos(t)}$

 

[Chipset3600]

Re: Integral- Without use integration technnique

How do you mean "without using integration techniques"? Surely you need integration techniques to solve an integral? Also what have you tried?

Hint: Let $u = \sec(t) = \frac{1}{\cos(t)}$

I mean without: substitution, integration by parts...

 

[Fantini]

Re: Integral- Without use integration technnique

I'm not sure that is possible, this is not an elementary integral (in the sense that it is the derivative of an elementary function, such as $\cos x$, $\sin x$, $\tan x$ and so on).

 

[soroban]

Re: Integral- Without use integration technnique

Hello, Chipset3600!

How can i solve this without use integration technniques?

. . \int \tan\theta \sec^3\!\theta\,d\theta

Well, maybe you can see all this?If we have: .f(x) \:=\:\tfrac{1}{3}\sec^3\!\theta + C

Then: .f'(x) \:=\:\tfrac{1}{3}\cdot 3\sec^2\!\theta\cdot\sec\theta\tan\theta + 0 \;=\;\tan\theta\sec^3\!\theta

 

[Chipset3600]

Re: Integral- Without use integration technnique

Hello, Chipset3600!


Well, maybe you can see all this?If we have: .f(x) \:=\:\tfrac{1}{3}\sec^3\!\theta + C

Then: .f'(x) \:=\:\tfrac{1}{3}\cdot 3\sec^2\!\theta\cdot\sec\theta\tan\theta + 0 \;=\;\tan\theta\sec^3\!\theta

\int tan(t).sec^3(t)dt = \int sec^2(t).sec(t).tan(t)dtUsing the power rule now:\frac{sec^3(t)}{3}+C

 

[Fantini]

Re: Integral- Without use integration technnique

You don't have a polynomial to apply the power rule, at least not until you use substitution.