General Q on integrating trig powers. (1 Viewer)

[nerdyasian]

I've been integrating most of the trig powers by myself, but sometimes i get different answers, if i don't do it by the method in the text book and get a different answer am i wrong?
for example, integrating: sec^4 ( x)
you can do it by parts, or split it up into trig identities, this will give you different answers?
So im wondering, is there a wrong approach? or are variations of the primitive accepted?
thanks.

also, wolfram alpha sometimes gives me a whole different answer as well.

 

[duckcowhybrid]

All methods should lead to the same answer. If it isn't, you are doing it wrong.

 

[mirakon]

I've been integrating most of the trig powers by myself, but sometimes i get different answers, if i don't do it by the method in the text book and get a different answer am i wrong?
for example, integrating: sec^4 ( x)
you can do it by parts, or split it up into trig identities, this will give you different answers?
So im wondering, is there a wrong approach? or are variations of the primitive accepted?
thanks.

also, wolfram alpha sometimes gives me a whole different answer as well.

are you sure it's not the same answer? Perhaps it's the same answer expressed differently e.g. you may not have simplified it enough. However if the answers are not equal then your solution is wrong, even different methods should provide the same solution.

 

[nerdyasian]

ahh ok, well after i got those replies, i double checked some of them, and they just need to be further simplified? but how am i suppose to know when to stop, because i can change the identities around a few times?

 

[duckcowhybrid]

Copy and Paste it into your post with Latex tags around it. Generally stop the moment you get gjdgjd + k, but if gjdgjd can be simplified easily, then do it.

 

[nerdyasian]

This is what i get:

in wolfram:

So, both ar right, and i just need to manipulate mine?

 

[Drongoski]

This is what i get:

in wolfram:

So, both ar right, and i just need to manipulate mine?

Your answer is right and need no further fiddling. Maybe Wolfram's is quoted with a term 'x' missing.

 

[nerdyasian]

ahhh ok thanks a bunch. one more question, integrating cosec cubed looks completely different to sec cubed, and i was wondering if any1 cud check this for me? seeing as though, wolfram sucks/ has variations.
my answer:


thanks all.

 

[Drongoski]

I could be wrong - but I think it is:

 

[Drongoski]

 

[nerdyasian]

lol thx dude, turns out i accidentally differentiated instead of integrated the csc in the 4th line... yet i differentiated wrong too. but yeah, thats right, and it looks like sec cubed too now, thanks. would rep you twice if i could.

 

[nikkifc]

Differentiate your answer to check.

Generally, to find the integral of odd powers of trig functions you require trig identities and for even powers require integration by parts I believe.