- #1

lmstaples

- 31

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## Homework Statement

Show that:

[itex]I = \int\int_{T}\frac{1}{(1 + x^{2})(1 + y^{2})}dxdy = \int^{1}_{0}\frac{arctan(x)}{(1 + x^{2})}dx = \frac{\pi^{2}}{32}[/itex]

where T is the triangle with successive vertices [itex](0,0), (1,0), (1,1)[/itex].

*By transforming to polar coordinates [itex](r,θ)[/itex] show that:*

[itex]I = \int^{\frac{\pi}{4}}_{0}\frac{log(\sqrt{2}cos(θ))}{cos(2θ)}dθ[/itex]

## Homework Equations

## The Attempt at a Solution

Part one is fine:

[itex]I = \int\int_{T}\frac{1}{(1 + x^{2})(1 + y^{2})}dxdy = \int^{1}_{0}\underbrace{\int^{x}_{0}\frac{1}{(1 + x^{2})(1 + y^{2})}dy}_{A}dx[/itex]

[itex]A: \frac{1}{(1 + x^{2})}\int^{x}_{0}\frac{1}{(1 + y^{2})}dy = \frac{1}{(1 + x^{2})}\left[arctan(y)\right]^{x}_{0} = \frac{arctan(x)}{(1 + x^{2}}[/itex]

[itex]\Rightarrow I = \int^{1}_{0}\frac{arctan(x)}{(1 + x^{2})}dx[/itex]

[itex]Let:[/itex] [itex]u = arctan(x)[/itex]

[itex]\frac{du}{dx} = \frac{1}{(1 + x^{2})} \Rightarrow du = \frac{1}{(1 + x^{2})}dx[/itex]

[itex]x = 0 \Rightarrow u = 0[/itex]

[itex]x = 1 \Rightarrow u = \frac{\pi}{4}[/itex]

[itex]\Rightarrow I = \int^{\frac{\pi}{4}}_{0}udu = \left[\frac{1}{2}u^{2}\right]^{\frac{\pi}{4}}_{0} = \frac{\pi^{2}}{32}.[/itex]

PART 2

[itex]x = rcos(\vartheta)[/itex]

[itex]y = rsin(\vartheta)[/itex]

[itex]Boundaries:[/itex]

[itex](i) y = 0 \Rightarrow rsin(θ) = 0[/itex]

[itex](ii) x = 1 \Rightarrow rcos(θ) = 1[/itex]

[itex](iii) y = x \Rightarrow rsin(θ) = rcos(θ) \Rightarrow tan(θ) = 1 \Rightarrow θ = \frac{\pi}{4}[/itex]

From then on everything seems to get very complex and I'm not really sure what the limits are either.

I was wondering if it was possible to go simply from:

[itex]I = \int^{1}_{0}\frac{arctan(x)}{(1 + x^{2})}dx[/itex]

to:

[itex]I = \int^{\frac{\pi}{4}}_{0}\frac{log(\sqrt{2}cos(θ))}{cos(2θ}dθ[/itex]

Please help :(