BITSAT 2019 Paper

(a) Let I=

π 2

∫

0

xsin‌2x cos‌2x dx ...(i)
From the definite integral property

a

∫

0

f (x) dx =

a

∫

0

f (a - x) dx
we have
I=

π 2

∫

0

(

π

2

−x)sin‌2x cos‌2x dx ...(ii)
(∵cos2x=sin2(

π

2

−x) & sin2x=cos2(

π

2

−x))
By adding (i) and (ii)
2I=

π

2

π 2

∫

0

sin‌2x cos‌2x dx
or 2I=

π

8

π 2

∫

0

sin‌22x dx
[∵ sin 2x = 2 sin x cos x]
=

π

8

π 2

∫

0

(1 - cos4x)dx (∵ cos2θ = 1 - 2sin‌2θ)
⇒ 2I=

π

8

[x−

sin‌4‌x

4

]0

π

2

⇒ 2I=

π

8

[

π

2

−0] ⇒ I=

π2

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