Integral with substitution and parts

**Steelers72** · March 9th 2013, 03:45 PM

∫ sin(2t)*e^[cos(t)] dt (from t=0 to π)

My work:
trig identity: sin2t= 2sintcost
bring constant to the outside
=> 2 ∫ sin(t)cos(t)*e^[cos(t)] dt (from t=0 to π)
I did u substitution and did
u=cos(t)
du=-sin(t)dt

new intervals:
cos(pi)=-1
cos(0)=1

Im not sure what to after this. Do I divide du=-sint(dt) by negative 1 and bring that -1 to the outside of integral since there is only a sin(t)dt not -sin(t)dt?

I appreciate any help!

**Steelers72** · March 9th 2013, 05:54 PM

Ok I figured out what to do, but this last algebra is bugging me.

= 2 * [(e - e) - (-1/e - 1/e)]
= 4/e

does it simplify to 0 (+2/e +2/e)

which equals 4/e?

**Soroban** · March 9th 2013, 07:48 PM

Hello, Steelers72!

I couldn't follow what you did, but you got the right answer.

$\int^{\pi}_0\sin(2t)e^{\cos t}\,dt$

I'll drop the limits for now . . .

We have: . $I \;=\;\int 2\sin t\cos t\,e^{\cos t}dt \;=\;2\int\cos t(\sin t\,e^{\cos t}dt)$

By parts: . $\begin{Bmatrix}u &=& \cos t && dv &=& \sin t\,e^{\cos t}dt \\ du &=& \text{-}\sin t\,dt && v &=& \text{-}e^{\cos t} \end{Bmatrix}$

Then: . $I \;=\;2\left[\text{-}e^{\cos t}\cos t - \int e^{\cos t}\sin t\,dt\right]$

. . . . . $I \;=\;2\left[\text{-}e^{\cos t}\cos t + \int e^{\cos t}(\text{-}\sin t\,dt)\right]$

. . . . . $I \;=\;2\left[\text{-}e^{\cos t}\cos t + e^{\cos t}\right] + C$

. . . . . $I \;=\;2e^{\cos t}(1-\cos t) + C$

Evaluate: . $2e^{\cos t}(1-\cos t)\,\bigg]^{\pi}_0$

. . . . . . . $=\;2e^{\cos\pi}(1-\cos\pi) - 2e^{\cos0}(1 - \cos 0)$

. . . . . . . $=\;2e^{-1}(1+1) - 2e^1(1-1)$

. . . . . . . $=\;\frac{4}{e}$

**Steelers72** · March 9th 2013, 10:10 PM

Thanks for the help! Just one question:

When you took the integral of e^cost (-sintdt) , why did the -sint get eliminated? The antiderivative of -sin is cos right? I'm confused at that part

**ibdutt** · March 9th 2013, 10:31 PM

Alternatively making a substitution can make the integral simple.
let cos t = u
-sint dt = du
thus we will get

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