The value of the integral ∫0121+3((x+1)2(1−x)6)14dx is

∫012(1+3)((x+1)2(1−x)6)14dx= ∫012(1+3)((1−x2)2(1−x)4)14dx =[∫0121(1−x2)(1−x)dx](1+3)put x = sin θ=[∫0π6cosθdθcosθ(1−sinθ)](1+3)=[∫0π61+sinθ dθcos2 θ](1+3)=[∫0π6sec2 θdθ+∫0π6tanθ secθdθ](1+3)=[ tanθ|0π6++ secθ|0π6](1+3)13+(23−1)(1+3)=(3−1)(3+1)=(3)2−1=2

The value of the integral ∫0121+3((x+1)2(1−x)6)14dx is

∫012(1+3)((x+1)2(1−x)6)14dx= ∫012(1+3)((1−x2)2(1−x)4)14dx =[∫0121(1−x2)(1−x)dx](1+3)put x = sin θ=[∫0π6cosθdθcosθ(1−sinθ)](1+3)=[∫0π61+sinθ dθcos2 θ](1+3)=[∫0π6sec2 θdθ+∫0π6tanθ secθdθ](1+3)=[ tanθ|0π6++ secθ|0π6](1+3)13+(23−1)(1+3)=(3−1)(3+1)=(3)2−1=2

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The value of the integral $\int_{0}^{\frac{1}{2}} \frac{1 + \sqrt{3}}{{({(x + 1)}^{2} {(1 - x)}^{6})}^{\frac{1}{4}}} d x$ is

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Detailed Solution

$\begin{array}{l} \int_{0}^{\frac{1}{2}} \frac{(1 + \sqrt{3})}{{({(x + 1)}^{2} {(1 - x)}^{6})}^{\frac{1}{4}}} d x = \int_{0}^{\frac{1}{2}} \begin{array}{l} \frac{(1 + \sqrt{3})}{{({(1 - x^{2})}^{2} {(1 - x)}^{4})}^{\frac{1}{4}}} d x \end{array} \\ = [\int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{(1 - x^{2})} (1 - x)} d x] (1 + \sqrt{3}) \\ p u t x = sin θ \\ = [\int_{0}^{\frac{π}{6}} \frac{\cos θ d θ}{\cos θ (1 - \sin θ)}] (1 + \sqrt{3}) \\ = [\int_{0}^{\frac{π}{6}} \frac{1 + \sin θ d θ}{\cos^{2} θ}] (1 + \sqrt{3}) \\ = [\int_{0}^{\frac{π}{6}} \sec^{2} θ d θ + \int_{0}^{\frac{π}{6}} \tan θ sec θ d θ] (1 + \sqrt{3}) \\ = [{\tan θ |}_{0}^{\frac{π}{6}} + + {\sec θ |}_{0}^{\frac{π}{6}}] (1 + \sqrt{3}) \\ \frac{1}{\sqrt{3}} + (\frac{2}{\sqrt{3}} - 1) (1 + \sqrt{3}) \\ = (\sqrt{3} - 1) (\sqrt{3} + 1) \\ = {(\sqrt{3})}^{2} - 1 = 2 \end{array}$