integration of trig (1 Viewer)

[izzy_xo]

hey just wondering if anyone could help me

Integrate cosx/1+2sinx

cheers.

 

[Tofuu]

i think (i'm not very confident without answers =.=)

 

[biopia]

Quite simple, my friend. However this one looks a little harder because of the trig involved!

Notice that the derivative of sin is cos and you should realise that the numerator is ALMOST the derivative of the denominator.

The derivative of 1 + 2sinx = 2cosx

Since we have half of this on the numerator, the answer is half the log of the denominator.
i.e.:
∫cosx/(1+2sinx) = 1/2.log(1+2sinx) + C

If you don't quite understand, just say so, and I'll try to explain it a bit better =]

Hope that helps!

 

[izzy_xo]

hey thanks for that, really was simple !

 

[Tofuu]

Quite simple, my friend. However this one looks a little harder because of the trig involved!

Notice that the derivative of sin is cos and you should realise that the numerator is ALMOST the derivative of the denominator.

The derivative of 1 + 2sinx = 2cosx

Since we have half of this on the numerator, the answer is half the log of the denominator.
i.e.:
∫cosx/(1+2sinx) = 1/2.log(1+2sinx) + C

If you don't quite understand, just say so, and I'll try to explain it a bit better =]

Hope that helps!

+1

i think the rule is also on the standard integrals table

remember, you can ln the denominator as long as there's no x's (including trig, exponentials...etc) in the denominator or numerator UNLESS its inside the ln thingo

eg.




you cannot integrate this by using ln because the result will have cosx left on the numerator








as mentioned above, you cannot integrate this by ln it directly becuase you will have an x on the denominator







so long as all other 'x's are removed, you should be able to use ln

 

[biopia]

+1

i think the rule is also on the standard integrals table

remember, you can ln the denominator as long as there's no x's (including trig, exponentials...etc) in the denominator or numerator UNLESS its inside the ln thingo

eg.




you cannot integrate this by using ln because the result will have cosx left on the numerator








as mentioned above, you cannot integrate this by ln it directly becuase you will have an x on the denominator







so long as all other 'x's are removed, you should be able to use ln

That first one isn't as simple as putting the extra cosx out the front :S
Check here: http://www.wolframalpha.com/input/?i=integrate+(cosx)^2/(1+2sinx)

It's quite complicated...
It's because you are integrating with respect to x, and cosx is something with an x, so you still have to integrate it.

The second one is also a little confusing and over simplified...
integrate x/(2+x^3) - Wolfram|Alpha

Again, you still have to take into consideration that extra x.

 

[Tofuu]

That first one isn't as simple as putting the extra cosx out the front :S
Check here: integrate (cosx)^2/(1+2sinx) - Wolfram|Alpha

It's quite complicated...
It's because you are integrating with respect to x, and cosx is something with an x, so you still have to integrate it.

The second one is also a little confusing and over simplified...
integrate x/(2+x^3) - Wolfram|Alpha

Again, you still have to take into consideration that extra x.

yea, i was just giving an example where/how putting ln will not work

the ans from Wolfarm===>LOL

 

[biopia]

yea, i was just giving an example where/how putting ln will not work

the ans from Wolfarm===>LOL

Oh ok. Sorry about that!

Yeah, I know aye... Those answers are INSANE! :S