The value of 8 π ∫ 0 π 2 cos x 2023 sin x 2023 + cos x 2023 d x is ______.

The value of 8π0π2cosx2023sinx2023+cosx2023dx is ______.

Solution

Let

I=8π0π2cosx2023sinx2023+cosx2023dx   ...1

Using 0af(x)dx=0af(a-x)dx

I=8π0π2sinx2023sinx2023+cosx2023dx   ...2

Adding 1 and 2, we get

2I=8π0π21 dx

2I=8πx0π2

I=2

Asked in: JEE Main 2023 (24 Jan Shift 1)

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