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

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- July 4th 2009, 03:35 AMsmartcar29indefinite 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 - July 4th 2009, 03:43 AMmalaygoel
- July 4th 2009, 03:54 AMsmartcar29
many thanks to you, of course its 1/u so needed to flip it. thanks again (Rofl)

- July 4th 2009, 04:20 AMtom@ballooncalculus
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...

http://www.ballooncalculus.org/asy/p...oubleAngle.png

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 - July 4th 2009, 04:35 AMsmartcar29
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