First make a substitution and then use integration by parts to evaluate the integral. \(\int_{0}^{π} e^{cost} sin 2t dt.\)

Solution:

Given: \(\int_{0}^{π} e^{cost} sin 2t dt.\)

\(\int_{0}^{π} e^{cost} sin 2t dt.\) = 2\(\int_{a}^{b} e^{cost} sin(t).cos(t)dt.\)----->

{we know that sin2t = 2sintcost}

Let us substitute u = cos t 

du = - sint dt

Thus \(\int_{0}^{π} e^{cost} sin 2t dt.\) = 2\(\int_{a}^{b} e^{cost} sin(t).cos(t)dt.\)

=  \(\int_{0}^{π} e^u . 2. u(-du)\)

=-2 \(\int_{0}^{π} e^u . u(du)\)

Now, integrate by parts: ∫ t.v = t.v - ∫vdt

Let

t = u

dt = 1. du

v = e^u

dv =e^u .du

-2 \(\int_{0}^{π} e^u . u(du)\)= t v - ∫vdt

= -2[ue^u -  ∫e^u .du]

= -2ue^u + 2 e^u

=[-2 cos t. e^{cos t} + 2 e^{cos t} ]\(_0^{\pi}\)

= (-2cosπe^cosπ + 2 e^cosπ) - (-2cos0e^cos0 + 2 e^cos0)

=(-2 (-1) e^-1 + 2 e^-1 )-(-2.1. e⁰ + 2e¹ )

= 2 e^-1 + 2e^-1 + 2 e - 2e

= 4 e^-1

= 4/e

therefore, \(\int_{0}^{π} e^{cost} sin 2t dt.\) = 4/e

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First make a substitution and then use integration by parts to evaluate the integral. \(\int_{a}^{b} e^cost sin 2t dt.\)

Summary:

By making a substitution and then using integration by parts to evaluate the integral, we got \(\int_{0}^{π} e^{cost} sin 2t dt.\) = 4/e.