1 Attachment(s)

Solving this double integral.

Hi.

I want to solve this.

Attachment 22915

I'm not sure what to do. I used integration by parts and I got stuck. I tried changing the order as well and i think it made it worse.

set:

u = 1/y

du = -1/y^2

dv = cos y

v = sin y

so i get

sin y/y + int (cos y)/y^2 from integration by parts.

If i change the order, I get the bounds as 0 to 1 for the outer integral and y to 1 for the inner integral and switching the dx and dy. That gives me

(cos y)/y - cos y for the inner integral.

So I'm not sure what to do now...

Re: Solving this double integral.

Quote:

Originally Posted by

**Kuma** Hi.

I want to solve this.

Attachment 22915
I'm not sure what to do. I used integration by parts and I got stuck. I tried changing the order as well and i think it made it worse.

set:

u = 1/y

du = -1/y^2

dv = cos y

v = sin y

so i get

sin y/y + int (cos y)/y^2 from integration by parts.

If i change the order, I get the bounds as 0 to 1 for the outer integral and y to 1 for the inner integral and switching the dx and dy. That gives me

(cos y)/y - cos y for the inner integral.

So I'm not sure what to do now...

The first thing you need to do is change the order of integration.

From your bounds, you can see $\displaystyle \displaystyle 0 \leq x \leq 1 $ and $\displaystyle \displaystyle x \leq y \leq 1$.

It should be clear therefore that $\displaystyle \displaystyle 0 \leq x \leq y$ and $\displaystyle \displaystyle 0 \leq y \leq 1$.

So your double integral becomes

$\displaystyle \displaystyle \begin{align*} \int_0^1{\int_x^1{\frac{\cos{y}}{y}\,dy}\,dx} &= \int_0^1{\int_0^y{\frac{\cos{y}}{y}\,dx}\,dy} \\ &= \int_0^1{\left[\frac{x\cos{y}}{y}\right]_0^y\,dy} \\ &= \int_0^1{\left(\frac{y\cos{y}}{y} - \frac{0\cos{y}}{y}\right)\,dy} \\ &= \int_0^1{\cos{y}\,dy} \\ &= \left[\sin{y}\right]_0^1 \\ &= \sin{1} - \sin{0} \\ &= \sin{1} \end{align*}$