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

Question

Luke Alvarez · Accepted Answer

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

#intf(x)g'(x)dx=f(x)g(x)-intg(x)f'(x)dx#.

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

#sin2alpha=2sinalphacosalpha#.

Thus, the integral takes on the following form:

#2int_0^pie^costsintcostdt=-2int_0^pie^cost(-sint)*costdt=(1)#

We can presume that:

#f(t)=cost# and #g'(t)dx=e^cost(-sint)dt#.

So:

#f'(t)dt=-sintdt#

#g(t)=e^cost# and this is because #inte^f(x)f'(x)dx=e^f(x)+c#.

Then:

#(1)=-2{[cost(e^cost)]_0^pi-int_0^pie^cost(-sint)dt}=#

#=-2[coste^cost-e^cost]_0^pi=#

#=-2(cospie^cospi-e^cospi-(cos0e^cos0-e^cos0))=#

#=-2(-e^-1-e^-1-e+e)=4/e#.

Aurora Alvarado · Answer

undefined 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.