# integrationWatch

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Thread starter 13 years ago
#1
hey
just having a bit of a blank,
how would you integrate

(cost)^4+(sint)^4
0
13 years ago
#2
(Original post by 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
0
13 years ago
#3
Just remember:

0
13 years ago
#4
(Original post by 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
0
13 years ago
#5
(Original post by 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
arghh. I had that initially. Why did I change it??
0
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