integrate (x^3)(cosh2x)

Mar 2014
909
2
malaysia
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why the ans from the tabular method is different from the by part method ? why the ans from for each term of tabular method is more than the by part method by 1/2 ?
 
Mar 2014
909
2
malaysia
can someone help pls ? thanks in advance !
 
Mar 2014
166
95
England
Hello. :)

What exactly is the problem? The end results don't seem different to me.
 
Mar 2014
909
2
malaysia
Hello. :)

What exactly is the problem? The end results don't seem different to me.
the first terms of the ans from the tabular method is(x^3)(cosh 2 x) , but not 1/2(x^3)(sinh 2 x) ... and so on....
but the final ans below the tabular method there give 1/2(x^3)(sinh 2 x) , tat's weird
 
Last edited:
Mar 2014
166
95
England
Remember that we move diagonally with the tabular approach - not directly across.

First, we have the positive term given by the product of \(\displaystyle x^3\) and \(\displaystyle \frac{1}{2}\sinh(2x)\) .

Then, we have the negative term given by the product of \(\displaystyle -3x^2\) and \(\displaystyle \frac{1}{4}\cosh(2x)\) .

...And so forth.
 
Last edited:
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