Definite integral

Calculus Level 5

If ∫ 0 1 d x ( 1 + x ) ( 2 + x ) x − x 2 = I \displaystyle \int_{0}^{1} \frac{dx}{(1+x)(2+x) \sqrt{x-x^2}} = I , then find 1000 I 1000I to the nearest integer.


The answer is 939.

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Jatin Yadav by jatin yadav , 23, India

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1 solution

Anish Puthuraya
Feb 10, 2014

Rearrange and simplify the integral to get,
∫ 0 1 2 d x ( 1 + x ) ( 2 + x ) 1 − ( 2 x − 1 ) 2 \displaystyle\int\limits_0^1 \frac{2dx}{(1+x)(2+x)\sqrt{1-(2x-1)^2}}

Substituting 2 x − 1 = sin ⁡ θ \displaystyle 2x-1 = \sin\theta ,
∫ − π / 2 π / 2 4 d θ ( 3 + sin ⁡ θ ) ( 5 + sin ⁡ θ ) = 2 ∫ − π / 2 π / 2 1 3 + sin ⁡ θ − 1 5 + sin ⁡ θ d θ \displaystyle\int\limits_{-\pi/2}^{\pi/2} \frac{4d\theta}{(3+\sin\theta)(5+\sin\theta)} = 2\int\limits_{-\pi/2}^{\pi/2} \frac{1}{3+\sin\theta} - \frac{1}{5+\sin\theta} d\theta

2 ∫ − π / 2 π / 2 1 3 + sin ⁡ θ d θ − 2 ∫ − π / 2 π / 2 1 5 + sin ⁡ θ d θ 2\displaystyle\int\limits_{-\pi/2}^{\pi/2} \frac{1}{3+\sin\theta}d\theta - 2\int\limits_{-\pi/2}^{\pi/2} \frac{1}{5+\sin\theta}d\theta

To evaluate both of these integrals, we use similar methods.
Substitute u = tan ⁡ ( θ / 2 ) \displaystyle u = \tan(\theta/2)

Hence,
sin ⁡ θ = 2 u 1 + u 2 \displaystyle \sin\theta = \frac{2u}{1+u^2} and cos ⁡ θ = 1 − u 2 1 + u 2 \displaystyle \cos\theta = \frac{1-u^2}{1+u^2}
Also,
d θ = 2 d u 1 + u 2 \displaystyle d\theta = \frac{2du}{1+u^2}

Using all these substitutions, we get,
4 ∫ − 1 1 d u 3 u 2 + 2 u + 3 − 4 ∫ − 1 1 d u 5 u 2 + 2 u + 5 \displaystyle 4\int\limits_{-1}^{1} \frac{du}{3u^2+2u+3} - 4\int\limits_{-1}^{1} \frac{du}{5u^2+2u+5}

Completing the square, integrating, and substituting the limits, we finally get,
I = 4 ( 0.55536 − 0.32063 ) = 0.93889 \displaystyle I = 4(0.55536 - 0.32063) = 0.93889

Hence,
1000 I = 938.89 ≈ 939 \displaystyle 1000I = 938.89 \approx \boxed{939}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

One could also have used definite integral properties:

∫ − π 2 π 2 d θ 3 + sin ⁡ θ = ∫ 0 π 2 ( 1 3 + sin ⁡ θ + 1 3 − sin ⁡ θ ) d θ \displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{d \theta}{3 + \sin \theta} = \int_{0}^{\frac{\pi}{2}} \bigg(\frac{1}{3 + \sin \theta} + \frac{1}{3 - \sin \theta} \bigg) d \theta

= 6 ∫ 0 π 2 d θ 9 − sin ⁡ 2 θ \displaystyle 6 \int_{0}^{\frac{\pi}{2}} \frac{d \theta}{9 - \sin^2 \theta}

= ∫ 0 π 2 6 sec ⁡ 2 θ d θ 8 tan ⁡ 2 θ + 9 \displaystyle \int_{0}^{\frac{\pi}{2}} 6 \frac{\sec^2 \theta d \theta}{8 \tan^2 \theta + 9}

= 6 ∫ 0 ∞ d t 8 t 2 + 9 = π 2 2 \displaystyle 6 \int_{0}^{\infty} \frac{dt}{8t^2 + 9} = \frac{\pi}{2 \sqrt{2}}

Similarly, ∫ − π 2 π 2 d θ 5 + sin ⁡ θ = π 2 6 \displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{d \theta}{5 + \sin \theta} = \frac{\pi}{2 \sqrt{6}}

Hence, answer is 2 ( π 2 ( 1 2 − 1 6 ) ) = 0.93889 \displaystyle 2 \bigg(\frac{\pi}{2}\bigg(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{6}}\bigg)\bigg) = 0.93889

jatin yadav - 7 years, 4 months ago

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Yeah. That's a better approach.

Anish Puthuraya - 7 years, 4 months ago

good problem

and good solution

Anirudha Nayak - 7 years, 3 months ago

awssm soln....

Richik Mukhopadhyay - 7 years, 1 month ago

I have used exactly the same approach

Ronak Agarwal - 6 years, 10 months ago

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