Mathematics

The value of $$\displaystyle\int _{ 0 }^{ \pi /2 }{ \sin ^{ 6 }{ x }  } dx $$ is


ANSWER

$$\cfrac{5\pi}{32}$$


SOLUTION
$$\displaystyle \int _{0}^{\pi /2} \sin ^6 x\ dx$$

$$I=\displaystyle \int _{0}^{\pi /2} \cos ^6 x\ dx$$

$$2I=\displaystyle \int _{0}^{\pi /2} (\sin ^6 x+\cos^6 x) dx$$

$$2I=\displaystyle \int _{0}^{\pi /2} (\sin^2 x)^3+(\cos^2 x)^3 dx =\displaystyle \int _{0}^{\pi /2} (\sin^2 x+\cos^2x)(\sin^4 x+\cos^4x-\sin^2\cos^2x) dx$$

$$2I=\displaystyle \int _{0}^{\pi /2} (\sin^2 x+\cos^2x)^2-(3\sin^2 x\cos^2x) dx $$

$$2I=\displaystyle \int _{0}^{\pi /2} (1-3\sin^2 x\cos^2 x)dx\quad \cos 2x=2\cos^2x-1 =1-2\sin^2x$$

$$2I=\displaystyle \int _{0}^{\pi /2} \left (1-\dfrac {3}{4} \sin^2x \right)dx$$

$$2I=\displaystyle \int _{0}^{\pi /2} \left (1-\dfrac {3}{4} \left (1-\dfrac {\cos 4x}{2}\right)\right)=\displaystyle \int _{0}^{\pi /2} \left (1-\dfrac {3}{8} +\dfrac {3}{8} \cos x\right)dx$$

$$2I=\left [\dfrac {5}{8}x+\dfrac {3}{8} \dfrac {\sin 4x}{4}\right]_0^{\pi /2}=\dfrac {5}{7}\dfrac {\pi}{2}$$

$$2I=\dfrac {5\pi}{16}\ \Rightarrow \ I=\dfrac {5\pi}{32}$$

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Single Correct Medium Published on 17th 09, 2020
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