C4 Parametric Integration

13

Curve C has parametric equations x = sint and y = sin2t where 0 is < or equal to t < or equal to pie/2

a) Find the are of the region bounded by C and the x axis.

Using y.dx/dt i get sin2t.cost

Then i make that 2sintcost.cost
= 2sintcos^2t
= 2sint(1 - sin^2t)
= 2sint - 2sin^3t

Then i get stuck here. Integral of 2sin^3t?

Will give positive rep.
Regards, Asad

Reply 1

JohnSPals

asadtamimi

Curve C has parametric equations x = sint and y = sin2t where 0 is < or equal to t < or equal to pie/2

a) Find the are of the region bounded by C and the x axis.

Using y.dx/dt i get sin2t.cost

Then i make that 2sintcost.cost
= 2sintcos^2t
= 2sint(1 - sin^2t)
= 2sint - 2sin^3t

Then i get stuck here. Integral of 2sin^3t?

Will give positive rep.
Regards, Asad

There's the identity:
sin 3x = 3 sin x - 4 sin^3 x
that you could consider, perhaps?

2sin t - 2sin^3 t
= 2sin t - ((3/2) sin t - (1/2) sin 3t)
= 2sin t - (3/2) sin x + (1/2) sin 3x

OR, if you want to look at the integral in another way:

2sint - 2sin^3t
= 2sint [1-sin^2 t]
= 2sint.cos^2 t

Then use the change of variable u=cos t:

INT 2sint.cos^2 t dt
= INT -2.u^2 du

I'm sure one of these methods will be clear enough for you to understand.

Reply 2

Bape

2

asadtamimi

Curve C has parametric equations x = sint and y = sin2t where 0 is < or equal to t < or equal to pie/2

a) Find the are of the region bounded by C and the x axis.

Using y.dx/dt i get sin2t.cost

Then i make that 2sintcost.cost
= 2sintcos^2t
= 2sint(1 - sin^2t)
= 2sint - 2sin^3t

Then i get stuck here. Integral of 2sin^3t?

Will give positive rep.
Regards, Asad

Once you get

Int 2sintcos^2t dt

let y = cos^3t so dy/dt = 3 x cos^2t x -sint

therefore I = -(2/3)cos^3t

The sub in lims

C4 Parametric Integration

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