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Question

If  $$I_{1}=\displaystyle \int_{0}^{\pi/2}\sin^{4}x\ dx$$
$$I_{2}=\displaystyle \int_{0}^{\pi/2}\cos^{6}x\ dx$$
$$I_{3}=\displaystyle \int_{0}^{\pi/2}\sin^{8}x\ dx$$
$$I_{4}=\displaystyle \int_{0}^{\pi/2}\cos^{2}x\ dx$$ 
then the increasing order of $$I_{1},I_{2},I_{3},I_{4}$$  is?


A
I4,I2,I3,I1
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B
I3,I2,I1,I4
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C
I4,I3,I2,I1
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D
I3,I1,I2,I4
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Solution

The correct option is B $$I_{3},I_{2},I_{1},I_{4}$$

$$I_{1}=\int_{0}^{\pi/2} sinx^{4}dx ; I_{2}\int_{0}^{\pi/2}cos^{6}x dx; I_{3}=\int_{0}^{\pi/2}sinx^{8dx}$$
$$I_{4}\int_{0}^{\pi/2}cosx^{2}dx$$
Increasing order.
$$I_{1}=\dfrac{3}{4}\times\dfrac{1}{2}\times\pi/2; I_{2}=\dfrac{5}{6}\times\dfrac{3}{4}\times\dfrac{1}2\times\pi/2$$
$$I_{3}=\dfrac{7}{8}\times\dfrac{5}{6}\times\dfrac{3}{4}\times\dfrac{1}{2}\times\pi/2$$
$$I_{4}=\dfrac{1}{2}\times\pi/2$$So, increasing order
$$I_{3}, I_{2},I_{1},I_{4}$$


Mathematics

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