Question

integrate cosh 2x sinh 3x dx
can i solve this using by part?

Answer 1

I would write everything in terms of x.

Answer 2

\(\color{#000000 }{ \displaystyle \int \left(\frac{e^{2x}+e^{-2x}}{2}\right)\left(\frac{e^{3x}-e^{-3x}}{2}\right)dx}\)

Answer 3

why do we need to use definition?

Answer 4

\(\color{#000000 }{ \displaystyle \frac{1}{4}\int \left(e^{2x}+e^{-2x}\right)\left(e^{3x}-e^{-3x}\right)dx}\)

\(\color{#000000 }{ \displaystyle \frac{1}{4}\int \left(e^{5x}-e^{-x}+e^x-e^{-5x} \right)dx}\)

Answer 5

that should be simple enough to integrate \(\small \bf ....\)

Answer 6

I think that in this case it is the best, if not the best way.

Answer 7

i see, what other condition that we need to used definition?

Answer 8

can you rephrase that pleae, because I don't really understand what you want to know...

Answer 9

in this question, we integrate its definition. so, what type of question that is appropriate to integrate its definition.

Answer 10

we are (rather) integrating the function USING the definition.

We know that,
\(\color{#000000 }{ \displaystyle \frac{d}{dx} \cosh(x)=\sinh(x) }\)\(\tiny{\\[3.1em]}\)
\(\color{#000000 }{ \displaystyle \frac{d}{dx} \sinh(x)=\cosh(x) }\)

And these also come from the definition of hyperbolic functions, and by definition I am refering to how they are written in terms of x....

Answer 11

You can do the integration by parts perhaps
(with some u-substitution possibly)

Answer 12

i see, thank you

Answer 13

Although, when it is easier to use the "definition", then do that ... :)

Answer 14

YW