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integration

realicetic

hey
just having a bit of a blank,
how would you integrate

(cost)^4+(sint)^4

Reply 1

Christophicus

realicetic

hey
just having a bit of a blank,
how would you integrate

(cost)^4+(sint)^4

(cost)^4 + (sint)^4
= cos^4t + sin^4t
= cos^2t(cos^2t) + sin^2t(sin^2t)
= cos^2t(1-sin^2t) + sin^2t(1-cos^2t)
= cos^2t + sin^2t - 2sin^2t.cos^2t
= 1 - 2sin^2t.cos^2t
= 1 - 2sint.cost.sint.cost
= 1 - 0.5sin^2(2t)

cos4t= 1-2sin^2(2t)
=> sin^2(2t) = 0.5 - 0.5cos4t

1 - 0.5sin^2(2t)
= 1-0.5(0.5-0.5cos4t)
= 1-0.25+0.25cos4t
= 0.75+0.25cos4t

Reply 2

Kolya

Just remember:

\huge cos2x = cos^2x - sin^2x = 2cos^2x -1 = 1 - 2sin^2x

Reply 3

yazan_l

Widowmaker

= 1 - 2sint.cost.sint.cost
= 1 - 0.5sin^2(4t)

u have a mistake here:
it's :

1 - 2sint.cost.sint.cost
= 1 - (sin2t . ½sin2t)
= 1 - ½sin²2t

Edit: hence u will get to 0.75+0.25cos4t which integrates to 0.75t+1/16 sin4t

Reply 4

Christophicus

yazan_l

u have a mistake here:
it's :

1 - 2sint.cost.sint.cost
= 1 - (sin2t . ½sin2t)
= 1 - ½sin²2t

Edit: hence u will get to 0.75+0.25cos4t which integrates to 0.75t+1/16 sin4t