Chipset3600
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Hello MHB, how can i solve this without use integration technniques...
\int tan(t)sec^3(t)dt
\int tan(t)sec^3(t)dt
The discussion centers on solving the integral \(\int \tan(t) \sec^3(t) dt\) without employing traditional integration techniques such as substitution or integration by parts. Participants highlight that this integral is not elementary, meaning it cannot be expressed in terms of basic functions like sine or cosine. A hint is provided suggesting the use of the substitution \(u = \sec(t)\), which leads to a potential solution involving the expression \(\frac{1}{3} \sec^3(t) + C\). However, the consensus is that some form of integration technique is necessary to arrive at a solution.
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Chipset3600 said:Hello MHB, how can i solve this without use integration technniques...
\int tan(t)sec^3(t)dt
SuperSonic4 said:How do you mean "without using integration techniques"? Surely you need integration techniques to solve an integral? Also what have you tried?
Hint: Let $u = \sec(t) = \frac{1}{\cos(t)}$
Well, maybe you can see all this?If we have: .f(x) \:=\:\tfrac{1}{3}\sec^3\!\theta + CHow can i solve this without use integration technniques?
. . \int \tan\theta \sec^3\!\theta\,d\theta
soroban said: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