indefinite integrals

smartcar29 · #1 $Old$ July 4th, 2009, 04:35 AM

hi

can some one check this please

i have h(u)=sin^2 (3/4u)

i used sin^2 u = 1/2(1-cos(2u))

i made p=3/4u and sin^2 (p) = 1/2(1-cos(2p))

=1/2(1-cos(3/2u))

=1/2-1/2cos(3/2u))

integate=1/2u-3/4sin(3/2u)+c

many thanks

wayne

malaygoel · #2 $Old$ July 4th, 2009, 04:43 AM

Quote:

Originally Posted by smartcar29 $View Post$

hi

can some one check this please

i have h(u)=sin^2 (3/4u)

i used sin^2 u = 1/2(1-cos(2u))

i made p=3/4u and sin^2 (p) = 1/2(1-cos(2p))

=1/2(1-cos(3/2u))

=1/2-1/2cos(3/2u))....correct till here

integate=1/2u-3/4sin(3/2u)+c

many thanks

wayne

your answer is incorrect...

$\int (\frac{1}{2} - \frac{1}{2}\cos{\frac{3u}{2}})du=\frac{1u}{2} - \frac{1}{3}\sin{\frac{3u}{2}}+C$

smartcar29 · #3 $Old$ July 4th, 2009, 04:54 AM

many thanks to you, of course its 1/u so needed to flip it. thanks again

tom@ballooncalculus · #4 $Old$ July 4th, 2009, 05:20 AM

Quote:

Originally Posted by smartcar29 $View Post$

many thanks to you, of course its 1/u so needed to flip it. thanks again

Are you sure you got the point, here? 'Flipping' is involved, maybe, but in order to cancel the three-over-two fraction that comes as the derivative of 'three over two u', which is the inner function of a chain rule process.

Maybe you did entirely get the point - then forgive me this diagram opportunity...

As usual, straight continuous lines differentiate downwards (/integrate up), the straight dashed line similarly but with respect to the dashed balloon expression - so that the triangular network on the right satisfies the chain rule for differentiation.

The 'flipping' would be, perhaps, the multiplying of the one-over-two fraction by the reciprocal of three-over-two.

Don't integrate - balloontegrate!

Balloon Calculus Forum

smartcar29 · #5 $Old$ July 4th, 2009, 05:35 AM

thanks for the cool diagram, to be honest i do not fully understand but i have printed it off and will give it some careful study. many thanks for taking the time.

wayne