Ask your own question, for FREE!
Mathematics 7 Online
OpenStudy (anonymous):

calculus help

OpenStudy (bahrom7893):

calculus is here!

OpenStudy (anonymous):

Post the problem!!

OpenStudy (anonymous):

\[\int\limits_{?}^{?}(\cos ^{2}x-\cos ^{2}x)/(\sin x)dx\]

OpenStudy (anonymous):

opps the first cos is suppouse to be a sin

OpenStudy (bahrom7893):

Cos^2 - Sin^2?

OpenStudy (bahrom7893):

oh lol sorry it's the other way around

OpenStudy (anonymous):

\[\int\limits_{?}^{?}(\sin ^{2}x-\cos ^{2}x)/(\sin x)\]

OpenStudy (anonymous):

thats the problem

OpenStudy (anonymous):

You already asked this question. :| I didn't answer it, but you didn't reply to the old thread.

OpenStudy (anonymous):

i was waiting on you

OpenStudy (anonymous):

@badreferences

OpenStudy (anonymous):

I was giving you hint-by-hint. I realized just giving people answers would have them not learn anything.

OpenStudy (bahrom7893):

Just divide by sin Sinx - Cosx*Cotx

OpenStudy (anonymous):

Exactly what I said before. >.> Also, cosxcotx can be expressed as cscx-sinx.

OpenStudy (bahrom7893):

all right continue badrefs, finish ur answer!

OpenStudy (anonymous):

i am getting stuck on how you integrate that

OpenStudy (anonymous):

The integral of sinx is -cosx, riiight? The integral of cscx is -ln(sin(x/2))-ln(cos(x/2)). TBH, though, I forget how you prove that last part. It's just something I committed to memory and WolframAlpha lol.

OpenStudy (anonymous):

But like trig integrals, it's just something I remembered. So...

OpenStudy (anonymous):

ok thanks dude

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!