Calculus II Integration by Parts
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# Calculus II Integration by Parts

[From: ] [author: ] [Date: 11-09-17] [Hit: ]
= (1/2) [(1/(π + 2) + 1/2) - A].I hope this helps!......
If anyone can answer this, I'll give you a pat on the back and 10 points.
"Let A denote the value of the integral from 0 to pi (cosx / [(x+2)^2)])dx. Compute the following integral in terms of A: integral from 0 to pi/2 [(sinx*cosx) / (x+1)]dx."

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Note that
∫(x = 0 to π/2) (sin x cos x dx) / (x + 1)
= ∫(x = 0 to π/2) (1/2) sin(2x) dx / (x + 1)

Now, let w = 2x, dw = 2 dx.
Bounds: x = 0 ==> w = 0, and x = π/2 ==> w = π.

So, the integral becomes
∫(w = 0 to π) (1/2) sin w * (dw/2) / (w/2 + 1)
= ∫(w = 0 to π) (1/2) sin w dw / (w + 2)
= (1/2) ∫(x = 0 to π) sin x dx / (x + 2), dummy variable change

Finally, use integration by parts with
u = 1/(x + 2), dv = sin x dx
du = -1/(x + 2)^2, v = -cos x

We obtain
(1/2) [-cos(x)/(x + 2) {for x = 0 to π} - ∫(x = 0 to π) cos x dx / (x + 2)^2]
= (1/2) [(1/(π + 2) + 1/2) - A].

I hope this helps!
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keywords: Integration,by,Parts,II,Calculus,Calculus II Integration by Parts
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