1. Find the differential $dx$ and the new limits for $\theta$.
Using x = cos 2 θ x = \cos^2\theta x = cos 2 θ d x d θ = 2 cos θ ( − sin θ ) = − 2 cos θ sin θ \frac{dx}{d\theta} = 2\cos\theta(-\sin\theta) = -2\cos\theta\sin\theta d θ d x = 2 cos θ ( − sin θ ) = − 2 cos θ sin θ
Finding the new limits:x = 0 x = 0 x = 0 cos 2 θ = 0 ⟹ cos θ = 0 ⟹ θ = π 2 \cos^2\theta = 0 \implies \cos\theta = 0 \implies \theta = \frac{\pi}{2} cos 2 θ = 0 ⟹ cos θ = 0 ⟹ θ = 2 π x = 1 2 x = \frac{1}{2} x = 2 1 cos 2 θ = 1 2 ⟹ cos θ = 1 2 ⟹ θ = π 4 \cos^2\theta = \frac{1}{2} \implies \cos\theta = \frac{1}{\sqrt{2}} \implies \theta = \frac{\pi}{4} cos 2 θ = 2 1 ⟹ cos θ = 2 1 ⟹ θ = 4 π
2. Substitute $x$, $dx$, and the limits into the integral and simplify.
∫ 0 1 2 x 1 − x d x = ∫ π 2 π 4 cos 2 θ 1 − cos 2 θ ( − 2 cos θ sin θ ) d θ = ∫ π 2 π 4 cos θ sin 2 θ ( − 2 cos θ sin θ ) d θ = ∫ π 2 π 4 cos θ sin θ ( − 2 cos θ sin θ ) d θ = ∫ π 2 π 4 − 2 cos 2 θ d θ = ∫ π 4 π 2 2 cos 2 θ d θ \begin{aligned}
\int_{0}^{\frac{1}{2}} \frac{\sqrt{x}}{\sqrt{1-x}} \, dx &= \int_{\frac{\pi}{2}}^{\frac{\pi}{4}} \frac{\sqrt{\cos^2\theta}}{\sqrt{1-\cos^2\theta}} (-2\cos\theta\sin\theta) \, d\theta \\
&= \int_{\frac{\pi}{2}}^{\frac{\pi}{4}} \frac{\cos\theta}{\sqrt{\sin^2\theta}} (-2\cos\theta\sin\theta) \, d\theta \\
&= \int_{\frac{\pi}{2}}^{\frac{\pi}{4}} \frac{\cos\theta}{\sin\theta} (-2\cos\theta\sin\theta) \, d\theta \\
&= \int_{\frac{\pi}{2}}^{\frac{\pi}{4}} -2\cos^2\theta \, d\theta \\
&= \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} 2\cos^2\theta \, d\theta
\end{aligned} ∫ 0 2 1 1 − x x d x = ∫ 2 π 4 π 1 − cos 2 θ cos 2 θ ( − 2 cos θ sin θ ) d θ = ∫ 2 π 4 π sin 2 θ cos θ ( − 2 cos θ sin θ ) d θ = ∫ 2 π 4 π sin θ cos θ ( − 2 cos θ sin θ ) d θ = ∫ 2 π 4 π − 2 cos 2 θ d θ = ∫ 4 π 2 π 2 cos 2 θ d θ 3. Use the double angle identity to integrate.
Applying the identity 2 cos 2 θ = 1 + cos 2 θ 2\cos^2\theta = 1 + \cos 2\theta 2 cos 2 θ = 1 + cos 2 θ
∫ π 4 π 2 2 cos 2 θ d θ = ∫ π 4 π 2 ( 1 + cos 2 θ ) d θ = [ θ + 1 2 sin 2 θ ] π 4 π 2 \begin{aligned}
\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} 2\cos^2\theta \, d\theta &= \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (1 + \cos 2\theta) \, d\theta \\
&= \left[ \theta + \frac{1}{2}\sin 2\theta \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}
\end{aligned} ∫ 4 π 2 π 2 cos 2 θ d θ = ∫ 4 π 2 π ( 1 + cos 2 θ ) d θ = [ θ + 2 1 sin 2 θ ] 4 π 2 π 4. Substitute the limits to find the exact value.
[ θ + 1 2 sin 2 θ ] π 4 π 2 = ( π 2 + 1 2 sin ( π ) ) − ( π 4 + 1 2 sin ( π 2 ) ) = ( π 2 + 0 ) − ( π 4 + 1 2 ( 1 ) ) = π 4 − 1 2 \begin{aligned}
\left[ \theta + \frac{1}{2}\sin 2\theta \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} &= \left( \frac{\pi}{2} + \frac{1}{2}\sin(\pi) \right) - \left( \frac{\pi}{4} + \frac{1}{2}\sin\left(\frac{\pi}{2}\right) \right) \\
&= \left( \frac{\pi}{2} + 0 \right) - \left( \frac{\pi}{4} + \frac{1}{2}(1) \right) \\
&= \frac{\pi}{4} - \frac{1}{2}
\end{aligned} [ θ + 2 1 sin 2 θ ] 4 π 2 π = ( 2 π + 2 1 sin ( π ) ) − ( 4 π + 2 1 sin ( 2 π ) ) = ( 2 π + 0 ) − ( 4 π + 2 1 ( 1 ) ) = 4 π − 2 1 Answer π 4 − 1 2 \dfrac{\pi}{4} - \dfrac{1}{2} 4 π − 2 1