# Integration

• Mar 12th 2008, 05:42 PM
NeloAngelo
Integration
Hi I got a few probs with some integration questions such as

Integration by parts:

Limits [0,2] xsinh x dx
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Change of variable rule

Limits [-r,r] (r^2-x^2)^1/2 dx where x =rsint r>0
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and

Limits ["pie"/4, "pie"/2] (dx/sin x) using t = tan x/2)

Any kind of help is well appreciated

Thx
• Mar 12th 2008, 06:18 PM
topsquark
Quote:

Originally Posted by NeloAngelo
Hi I got a few probs with some integration questions such as

Integration by parts:

Limits [0,2] xsinh x dx

$\displaystyle \int _0^2 x~sinh(x)~dx$

Use integration by parts. u = x and v = sinh(x).

Quote:

Originally Posted by NeloAngelo
Change of variable rule

Limits [-r,r] (r^2-x^2)^1/2 dx where x =rsint r>0

$\displaystyle x = r~sin(t) \implies dx = r~cos(t)~dt$

So
$\displaystyle \int (r^2 - x^2)^{1/2}~dx = \int (r^2 - r^2~sin^2(t))^{1/2}~(r~cos(t)~dt)$

$\displaystyle = \int r^2 (1 - sin^2(t))^{1/2}~cos(t)~dt$

$\displaystyle = \int r^2 cos^2(t)~dt$

Now for the integration limits.
$\displaystyle x = -r \implies -r = r~sin(t) \implies sin(t) = -1 \implies t = -\frac{pi}{2}$

$\displaystyle x = r \implies -r = r~sin(t) \implies sin(t) = 1 \implies t = \frac{pi}{2}$

So the final form for the integral is
$\displaystyle r^2 \int_{-\pi/2}^{\pi/2} cos^2(t)~dt$

-Dan
• Mar 13th 2008, 10:29 AM
Krizalid
Quote:

Originally Posted by NeloAngelo
Change of variable rule

Limits [-r,r] (r^2-x^2)^1/2 dx where x =rsint r>0

It's a semicircle whose radius is $\displaystyle r,$ hence its value is $\displaystyle \frac\pi2r^2.$

Quote:

Originally Posted by topsquark
So the final form for the integral is
$\displaystyle r^2 \int_{-\pi/2}^{\pi/2} cos^2(t)~dt$

Having this, don't even think to replace the double angle formula for cosine.

First, note that this function is even, so the integral can be rewritten as $\displaystyle 2r^2\int_0^{\pi /2} {\cos ^2 t\,dt} .$ Besides $\displaystyle \int_0^{\pi /2} {f(\sin x)\,dx} = \int_0^{\pi /2} {f(\cos x)\,dx}$ whose proof is easy, so $\displaystyle \int_0^{\pi /2} {\cos ^2 t\,dt} = \int_0^{\pi /2} {\sin ^2 t\,dt} .$ Let $\displaystyle \lambda$ be the value of these integrals, so

$\displaystyle 2\lambda = \int_0^{\pi /2} {dt} \Rightarrow \lambda = \frac{\pi } {4},$ and the conclusion follows.
• Mar 13th 2008, 06:24 PM
NeloAngelo
Hi thx for feedback final problem I had was..

Limits [1,10] (1*x^-1/2)*log(x^1/2) dx where x = t^2
First thought that came to mind was the square roots would cancel when substituing t and dt = 1/2*t^3

but the log part left me fogged out only idea I can think of is "(log*x^1/2) ---> 1/t + Constant.

Anything else I could do?