Solvable, but..

Calculus Level 4

∫ 0 2 3 ( π 3 − arcsin ⁡ 4 x 2 + 1 − 1 2 x ) d x = ? \int_0^{2\sqrt3} \left(\frac{\pi}3-\arcsin{\frac{\sqrt{4x^2+1}-1}{2x}} \right) dx = \ ?

Try solving this without integrating arcsin.

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Inquisitor Math by Inquisitor Math , 20, South Korea

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2 solutions

Aryan Gupta
Nov 14, 2020

I used xdy+ydx=d(xy). sin y=that big func. And then x=tany×secy

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Yes, the key was that this integral was equivalent to ∫ 0 π 3 s i n x 1 − ( s i n x ) 2 d x = 1 \int_0^{\frac{\pi}3} \frac{sinx}{1-(sinx)^2} \, dx = 1 . (Was x = tan ⁡ y sec ⁡ y x= \tan{y} \sec{y} based on pure observation?)

Inquisitor Math - 6 months, 4 weeks ago
Inquisitor Math
Nov 14, 2020

Solvable because (according to wolfram) ∫ arcsin ⁡ 4 x 2 + 1 − 1 2 x d x = x arcsin ⁡ 4 x 2 + 1 − 1 2 x − 4 x 2 + 1 − 1 x 2 ( 4 x 2 + 1 + 1 ) 2 2 + c o n s t a n t \int \arcsin{\frac{\sqrt{4x^2+1}-1}{2x}} \, dx = x \arcsin{\frac{\sqrt{4x^2+1}-1}{2x}} - \frac{\sqrt\frac{\sqrt{4 x^2 + 1} - 1}{x^2} (\sqrt{4 x^2 + 1} + 1)}{2 \sqrt2} + constant but this problem can be approached differently.

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