# How do integrate #int e^(cos)(t) (sin 2t) dt# between a = 0, #b = pi#?

Now, by portion

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To integrate ( \int_{0}^{\pi} e^{\cos(t)} \sin(2t) , dt ) over the interval ( [0, \pi] ), you can use integration by parts. Let ( u = e^{\cos(t)} ) and ( dv = \sin(2t) , dt ). Then, ( du = -e^{\cos(t)} \sin(t) , dt ) and ( v = -\frac{1}{2}\cos(2t) ).

Applying the integration by parts formula ( \int u , dv = uv - \int v , du ), you'll get:

[ \int_{0}^{\pi} e^{\cos(t)} \sin(2t) , dt = \left[-\frac{1}{2} e^{\cos(t)} \cos(2t)\right]*{0}^{\pi} - \int*{0}^{\pi} \left(-\frac{1}{2}\cos(2t)\right) (-e^{\cos(t)} \sin(t)) , dt ]

[ = -\frac{1}{2} e^{\cos(\pi)} \cos(2\pi) + \frac{1}{2} e^{\cos(0)} \cos(0) - \frac{1}{2} \int_{0}^{\pi} e^{\cos(t)} \cos(2t) \sin(t) , dt ]

[ = -\frac{1}{2} e^{-1} + \frac{1}{2} - \frac{1}{2} \int_{0}^{\pi} e^{\cos(t)} \cos(2t) \sin(t) , dt ]

The remaining integral might need further manipulation or a different integration technique to solve.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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