Thread: indefinite integrals

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

Originally Posted by smartcar29

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...

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

Originally Posted by smartcar29

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

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