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Using the substitution x=sin^(2)theta, or otherwise, evaluate int_(0)^((1)/(2))sqrt((x)/(1-x))dx.

Using the substitution x=sin⁡2θx=\sin^{2}\theta, or otherwise, evaluate ∫012x1−xdx\int_{0}^{\frac{1}{2}}\sqrt{\frac{x}{1-x}}dx.

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Full solution

Q. Using the substitution x=sin⁡2θx=\sin^{2}\theta, or otherwise, evaluate ∫012x1−xdx\int_{0}^{\frac{1}{2}}\sqrt{\frac{x}{1-x}}dx.
  1. Make Substitution: Let's make the substitution x=sin⁡2(θ)x = \sin^2(\theta). Then, dxdx will be the derivative of sin⁡2(θ)\sin^2(\theta) with respect to θ\theta, which is 2sin⁡(θ)cos⁡(θ)d(θ)2\sin(\theta)\cos(\theta)d(\theta) or sin⁡(2θ)d(θ)\sin(2\theta)d(\theta).
  2. Change Limits: We need to change the limits of integration to match our substitution. When x=0x = 0, sin⁡2(θ)=0\sin^2(\theta) = 0, so θ=0\theta = 0. When x=12x = \frac{1}{2}, sin⁡2(θ)=12\sin^2(\theta) = \frac{1}{2}, so θ=π4\theta = \frac{\pi}{4}. Therefore, our new limits of integration are from 00 to π4\frac{\pi}{4}.
  3. Substitute and Simplify: Substitute xx with sin⁡2(θ)\sin^2(\theta) and dxdx with sin⁡(2θ)d(θ)\sin(2\theta)d(\theta) in the integral. The integral becomes:\newline∫0π/4sin⁡2(θ)1−sin⁡2(θ)⋅sin⁡(2θ)d(θ)\int_{0}^{\pi/4} \sqrt{\frac{\sin^2(\theta)}{1-\sin^2(\theta)}} \cdot \sin(2\theta)d(\theta)
  4. Use Trigonometric Identity: Simplify the integral. Since 1−sin⁡2(θ)1 - \sin^2(\theta) is cos⁡2(θ)\cos^2(\theta), the integral becomes:\newline∫0π/4sin⁡2(θ)cos⁡2(θ)⋅sin⁡(2θ) d(θ)\int_{0}^{\pi/4} \sqrt{\frac{\sin^2(\theta)}{\cos^2(\theta)}} \cdot \sin(2\theta)\,d(\theta)\newlineThis simplifies to:\newline∫0π/4tan⁡(θ)⋅sin⁡(2θ) d(θ)\int_{0}^{\pi/4} \tan(\theta) \cdot \sin(2\theta)\,d(\theta)
  5. Cancel Terms: We can use a trigonometric identity to simplify sin⁡(2θ)\sin(2\theta) to 2sin⁡(θ)cos⁡(θ)2\sin(\theta)\cos(\theta). The integral now becomes:\newline∫0π/4tan⁡(θ)⋅2sin⁡(θ)cos⁡(θ) d(θ)\int_{0}^{\pi/4} \tan(\theta) \cdot 2\sin(\theta)\cos(\theta)\,d(\theta)
  6. Apply Double-Angle Identity: Since tan⁡(θ)\tan(\theta) is sin⁡(θ)/cos⁡(θ)\sin(\theta)/\cos(\theta), we can cancel out the cos⁡(θ)\cos(\theta) in the denominator with one of the cos⁡(θ)\cos(\theta) in the numerator. The integral simplifies to:\newline∫0π/42sin⁡2(θ)d(θ)\int_{0}^{\pi/4} 2\sin^2(\theta)d(\theta)
  7. Split Integral: Use the double-angle identity for sine, which is sin⁡2(θ)=1−cos⁡(2θ)2\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}. The integral becomes:\newline∫0π42⋅1−cos⁡(2θ)2d(θ)\int_{0}^{\frac{\pi}{4}} 2 \cdot \frac{1 - \cos(2\theta)}{2} d(\theta)\newlineThis simplifies to:\newline∫0π4(1−cos⁡(2θ))d(θ)\int_{0}^{\frac{\pi}{4}} (1 - \cos(2\theta)) d(\theta)
  8. Integrate First Part: Split the integral into two parts and integrate each part separately: ∫0π/41 dθ−∫0π/4cos⁡(2θ) dθ\int_{0}^{\pi/4} 1 \, d\theta - \int_{0}^{\pi/4} \cos(2\theta) \, d\theta
  9. Integrate Second Part: Integrate the first part, which is the integral of 11 with respect to θ\theta from 00 to π/4\pi/4. This gives us:\newlineθ\theta_{00}^{π/4\pi/4} = π/4−0=π/4\pi/4 - 0 = \pi/4
  10. Combine Results: Integrate the second part, which is the integral of cos⁡(2θ)\cos(2\theta) with respect to θ\theta from 00 to π/4\pi/4. This gives us:\newline[12sin⁡(2θ)](0)(π/4)=12sin⁡(π2)−12sin⁡(0)=12−0=12[\frac{1}{2} \sin(2\theta)]_{(0)}^{(\pi/4)} = \frac{1}{2} \sin(\frac{\pi}{2}) - \frac{1}{2} \sin(0) = \frac{1}{2} - 0 = \frac{1}{2}
  11. Combine Results: Integrate the second part, which is the integral of cos⁡(2θ)\cos(2\theta) with respect to θ\theta from 00 to π/4\pi/4. This gives us:\newline[12sin⁡(2θ)](0)(π/4)=12sin⁡(π2)−12sin⁡(0)=12−0=12[\frac{1}{2} \sin(2\theta)]_{(0)}^{(\pi/4)} = \frac{1}{2} \sin(\frac{\pi}{2}) - \frac{1}{2} \sin(0) = \frac{1}{2} - 0 = \frac{1}{2}Combine the results of the two integrals to get the final answer. Subtract the result of the second integral from the first to get:\newlineπ/4−12\pi/4 - \frac{1}{2}
  12. Combine Results: Integrate the second part, which is the integral of cos⁡(2θ)\cos(2\theta) with respect to θ\theta from 00 to π/4\pi/4. This gives us:\newline[12sin⁡(2θ)](0)(π/4)=12sin⁡(π2)−12sin⁡(0)=12−0=12[\frac{1}{2} \sin(2\theta)]_{(0)}^{(\pi/4)} = \frac{1}{2} \sin(\frac{\pi}{2}) - \frac{1}{2} \sin(0) = \frac{1}{2} - 0 = \frac{1}{2}Combine the results of the two integrals to get the final answer. Subtract the result of the second integral from the first to get:\newlineπ/4−12\pi/4 - \frac{1}{2}Simplify the final answer. Since 12\frac{1}{2} is the same as 2/42/4, we can write:\newlineπ/4−2/4=(π−2)/4\pi/4 - 2/4 = (\pi - 2)/4

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