JEE Advanced 2020 Full Test 1 Paper 2

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Question : 49 of 54

Marks: +1, -0

Let I=

1∕2

∫

0

(‌

1

4

−x2)4dx. then

I=
1∕2
∫
0
x4(1−x)4dx
I=‌
1
2
‌
1
∫
0
x4(1−x)4dx
I=
π∕2
∫
0
sin9θcos9θdθ
I=‌
1
1260

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