Not so straightforward integral...

Calculus Level 3

Evaluate ∫ 2 x + 1 x − x sin ⁡ − 1 x 2 1 − x 2 d x \displaystyle \int \frac{2\sqrt{x} + \sqrt{\frac{1}{x} - x} \sin ^{-1}x} {2\sqrt{1-x^2}} dx .

If the above integral is of the form f ( x ) + C \displaystyle f(x) + C , then find f ( 1 ) f(1) . (round up to 2 decimals)

NOTE : f ( x ) f(x) is to be taken as the anti-derivative of the given integrand without adding any constant values. e.g. If the answer to the integral was sin ⁡ x + C \sin x +C , you take f ( x ) = sin ⁡ x f(x) = \sin x and NOT something like sin ⁡ x + 1 , sin ⁡ x + 2 \sin x + 1, \sin x + 2 , etc.


The answer is 1.57.

The Extremer by The Extremer , 19, India

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1 solution

Patrick Corn
Sep 25, 2019

This simplifies to ∫ x 1 − x 2 d x + ∫ sin ⁡ − 1 x 2 x d x . \int \frac{\sqrt{x}}{\sqrt{1-x^2}} \, dx + \int \frac{\sin^{-1} x}{2\sqrt{x}} \, dx. Integrating the second integral by parts with u = sin ⁡ − 1 x , d v = d x 2 x u = \sin^{-1} x, dv = \frac{dx}{2\sqrt{x}} gives ∫ x 1 − x 2 d x + x sin ⁡ − 1 x − ∫ x 1 − x 2 d x = x sin ⁡ − 1 x + C . \int \frac{\sqrt{x}}{\sqrt{1-x^2}} \, dx + \sqrt{x} \sin^{-1} x - \int \frac{\sqrt{x}}{\sqrt{1-x^2}} \, dx = \sqrt{x} \sin^{-1} x + C. The problem as stated is ambiguous, since I could have gotten x sin ⁡ − 1 x + 1 + C \sqrt{x} \sin^{-1} x + 1 + C as well, but apparently I was meant to plug in 1 1 to x sin ⁡ − 1 x , \sqrt{x} \sin^{-1} x, which gives π / 2 ≈ 1.57 . \pi/2 \approx \fbox{1.57}.

Suggestion: fix the problem by stipulating that f ( 0 ) = 0. f(0) = 0.

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Again, I think the NOTE is not quite sufficient. I mean, consider the well-known example: ∫ 2 sin ⁡ ( x ) cos ⁡ ( x ) d x = sin ⁡ 2 ( x ) + C = − cos ⁡ 2 ( x ) + C = − 1 2 cos ⁡ ( 2 x ) + C , \begin{aligned} \int 2 \sin(x) \cos(x) \, dx &= \sin^2(x) + C \\ &= -\cos^2(x) + C \\ &= -\frac12 \cos(2x) + C, \end{aligned} where each of those three functions doesn't have any "obvious" constant values added to it, although of course any pair of them differs by a constant. I still suggest you stipulate that f ( 0 ) = 0. f(0) = 0. Or, equivalently, change the integral to a definite integral, from 0 0 to 1. 1.

Patrick Corn - 1 year, 8 months ago

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