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Integrate [x f ⁻¹(x)] from 0 to √2, where f(x) = tan(x)sec(x)?
Just something I found fooling around. The answer is a nice number and there is a very simple (if round-about) way to evaluate it, but is it easy if you don't know that?
Correct answer Gianlino!
But how do we compute that exact value without going backwards, trying to compute something simple by going the hard way? Or is it obvious where I got this from??
My apologoies for confusing Gianlino. The problem arose from the area bounded by y=x and y=x² → very easily done: ∫(x-x²)dx. But I had been thinking lately about polar coordinates, so what would it look like integrated with respect to the angle θ? That led to (π/4 minus) this integral, which I thought looked rather difficult.
2 Answers
- IndicaLv 79 years agoFavorite Answer
Let y=f⁻¹(x) so x=f(y) and dx=f’(y)dy
∫ [x=0,√2] xf⁻¹(x)dx = ∫ [y=0,π/4] y{ f(y)f’(y)dy }
Integrating by parts gives ½yf²(y) – ½∫ f²(y)dy, [y=0,π/4]
With f(y)=sec(y)tan(y) this is ½ysec²(y)tan²(y) – (1/6)tan³(y), [y=0,π/4]
= π/4 – 1/6
- gianlinoLv 79 years ago
I would guess pi/4 - 1/6...
edit not sure I get what you mean with "getting backwards" or "the hard way"
