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Mathematics 23 Online
OpenStudy (anonymous):

what is the intergration of cos x

OpenStudy (anonymous):

sin x

OpenStudy (anonymous):

thanks

OpenStudy (jhannybean):

\[\large \int\limits \cos(x)dx = \sin(x) + \color{red}c\]

OpenStudy (anonymous):

actually tabkatta and Jhannybean. This lectures disagrees with u . says the intergral of cos x= -sin x. but the derivitve of cos x its -sin x. please watch the link http://www.mathtutor.ac.uk/integration/integrationbyparts/video

OpenStudy (jhannybean):

\[\large \frac{d}{dx}(\sin(x))= \cos(x) \therefore \int\limits \cos(x) = \sin(x) + c\]

OpenStudy (bahrom7893):

uhm integral of cosine is sine...

OpenStudy (bahrom7893):

integral of sine is negative cosine

OpenStudy (anonymous):

thanks a lot guys you are perfectly right

OpenStudy (anonymous):

and thanks for the link

OpenStudy (anonymous):

so what is the intergral of sine x?

OpenStudy (bahrom7893):

-Cos

OpenStudy (anonymous):

why - cos x?

OpenStudy (bahrom7893):

Because Derivative of Cos = -Sin Integral of Sin = Integral of -(-Sin) = Integral of - (Derivative of Cos) = - Integral of Derivative of Cosine = - Cosine

OpenStudy (bahrom7893):

if that makes any sense

OpenStudy (jhannybean):

\[\large \frac{d}{dx}(\cos(x)) = -\sin (x) \therefore -\int\limits \sin(x)dx= - \cos(x)\]

OpenStudy (jhannybean):

exactly what @bahrom7893 said, but in equation format.

OpenStudy (anonymous):

that. what is the intergral of -cox

OpenStudy (anonymous):

and -sin x

OpenStudy (jhannybean):

just take the negative outside the integral and integrate, add the negative in after.

OpenStudy (jhannybean):

I just did both of them for you

OpenStudy (anonymous):

so it means the intergration of -cos x its -sine x?

OpenStudy (anonymous):

thanks im right

OpenStudy (bahrom7893):

yea

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