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Question

Evaluate: $$\displaystyle \int_{0}^{\pi}\sin^{6}xdx$$


A
5π16
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B
35π128
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C
5π8
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D
5π18
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Solution

The correct option is A $$\displaystyle \frac{5\pi}{16}$$
$$I=\displaystyle\int_{0}^{\pi} \sin^6{x}dx$$

Now, If $$f(2a-x)=f(x)$$, then $$\displaystyle\int_{0}^{2a} f(x)dx=2\displaystyle\int_{0}^{a} f(x)dx$$

$$\implies I=2\displaystyle\int_{0}^{\pi/2} \sin^6{x}dx=I_1(say)$$

$$I=2\displaystyle\int_{0}^{\pi/2} \sin^6{\left(\dfrac{\pi}{2}-x\right)}dx$$

$$=2\displaystyle\int_{0}^{\pi/2} \cos^6{x}dx=I_2$$

Hence, $$2I=2\displaystyle\int_{0}^{\pi/2}\left(\sin^6{x}+\cos^6{x}\right)dx$$

$$\implies I=\displaystyle\int_{0}^{\pi/2} \left[(\sin^2{x}+\cos^2{x})^3-3\sin^2{x}\cos^2{x}(\sin^2{x}+\cos^2{x})\right]dx$$

$$\implies I=\displaystyle\int_{0}^{\pi/2} \left(1-3\sin^2{x}\cos^2{x}\right)dx$$

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

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

$$\implies I=\displaystyle\int_{0}^{\pi/2} \left(1-\dfrac{3}{4}.\dfrac{1-\cos{4x}}{2}\right)dx$$

$$\implies I=\displaystyle\int_{0}^{\pi/2} \left(\dfrac{5}{8}+\dfrac{3}{8}\cos{4x}\right)dx$$

$$\implies I=\dfrac{5}{8}\left[x\right]_{0}^{\pi/2}+\dfrac{3}{32}\left[\sin{4x}\right]_{0}^{\pi/2}$$

$$\implies I=\dfrac{5\pi}{16}$$

Hence, answer is option-(A).

Mathematics

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