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

The integration by parts theorem, which states the following, must be applied to this integral:

Additionally, we must keep in mind the sinus double-angle formula, which states:

Thus, the integral takes on the following form:

We can presume that:

So:

Then:

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To integrate ( \int_{0}^{\pi} e^{\cos(t)} \sin(2t) , dt ) between ( a = 0 ) and ( b = \pi ), you can use integration by parts.

Let ( u = e^{\cos(t)} ) and ( dv = \sin(2t) , dt ). Then, ( du = -\sin(t) e^{\cos(t)} , dt ) and ( v = -\frac{1}{2}\cos(2t) ).

Applying the integration by parts formula:

[ \int u , dv = uv \bigg|_{0}^{\pi} - \int v , du ]

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

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

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

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

Now, you can further integrate the remaining term or use numerical methods to approximate the integral value.

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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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