integral of t sin 2t dt

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Evaluate the integral below, where C is the curve r(t) = ‹sin(t), cos(t), sin(2t)›, 0 ≤ t ≤ 2π? Int (cos^2 t - sin t sin^2 2t + 2sin^3 t sin 2t)dt.
First makea substitution and then use integration by parts to evaluate. ∫e^(cos) (t) (sin 2t) dt. integrate equation 2 by - integration by parts.
integral(r in [0,1], t in [0, 2pi]) (8r^3 cos t sin^2(t) + 6r^3 cos^3(t) - 1) (r dt dr) = - integral(r in [0,1], t in [0, 2pi]) (8r^4 sin^2(t) cos t + 6r^4 (1.
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Evaluate the integral below, where C is the curve r(t) = ‹sin(t), cos(t), sin(2t)›, 0 ≤ t ≤ 2π? Int (cos^2 t - sin t sin^2 2t + 2sin^3 t sin 2t)dt.
First makea substitution and then use integration by parts to evaluate. ∫e^(cos) (t) (sin 2t) dt. integrate equation 2 by - integration by parts.

Evaluate the integral below, where C is the curver(t). More.

integral of t sin 2t dt

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Evaluate the integral below, where C is the curve r(t) = ‹sin(t.
Evaluate the integral below, where C is the curver(t) = ‹sin(t),cos(t), sin(2t)›, 0 ≤ t . integral(r in [0,1], t in [0, 2pi]) (8r^3 cos t sin^2(t) + 6r^3 cos^3(t) - 1) (r dt dr).
Let F(x) = ∫ (from 0 to x) sin(t^2 )dt for 0 ≤ x ≤ 3. On what.

integral of t sin 2t dt

How do I find f '(x) when f(x)= integral of sin^2t*dt from (x^2 to.
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evaluate the definite integral.. Best Answer - Chosen by Voters. π ∫ sin^2(t) * cos(t) dt 0 u = sin(t) du = cos(t) dt π ∫ u^2 * du 0..... π u^3/3 ].
I know you could use u substitution to say u = sin(t) then du = cos(t)dt, but I would still be left with e^(3t). Please help me solve this integral! this.
Proceed with integration by parts with u = t², du = 2t dt and dv = cos(t) dt, v = sin(t): t²sin(t) - 2 ∫ tsin(t) dt. Now do the same with the second.
in second integral substitute u=2t, du=2dt = (1/2) {from x² to 0} ∫ dt - (1/4) {from x² to 0} ∫ cos(u) du integral of 1 is t, integral of cos(u) is sin(u).
I will be using Int to mean the integral (since I don't know how to get the symbol. Add 1/4 Int( e^(-2t) sint dt) to both sides of the equation to get:.