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BITSAT 2019 Paper

Section: Mathematics

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Question : 97 of 150

Marks: +1, -0

\int\limits_{0}^{\pi/2}

^2

^2

x dx is equal to

$\frac{\pi^2}{32}$
$\frac{\pi^2}{16}$
$\frac{\pi}{32}$
None of these

Solution:

(a) Let

I = \int\limits_{0}^{\pi/2}

xsin

^2

x cos

^2

x dx ...(i)

From the definite integral property

\int\limits_{0}^{a}

f (x) dx =

\int\limits_{0}^{a}

f (a - x) dx

we have

I = \int\limits_{0}^{\pi/2} (\pi/2 - x)

sin

^2

x cos

^2

x dx ...(ii)

(\because \cos^2 x = \sin^2 (\pi/2 - x)

&

\sin^2 x = \cos^2 (\pi/2 - x))

By adding (i) and (ii)

2I = \pi/2 \int\limits_{0}^{\pi/2}

sin

^2

x cos

^2

x dx

or

2I = \pi/8 \int\limits_{0}^{\pi/2}

sin

^2

2x dx

[∵ sin 2x = 2 sin x cos x]

=

\pi/8 \int\limits_{0}^{\pi/2}

(1 - cos4x)dx (∵ cos2θ = 1 - 2sin

^2

θ)

⇒

2I = \pi/8

\left[ x - \frac{\sin 4x}{4} \right]_0^{\pi/2}

⇒

2I = \pi/8

[\pi/2 - 0]

⇒

I = \frac{\pi^2}{32}

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