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JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Definite Integrals

Let

f_{n} = \int_{0}^{\frac{π}{2}} (\sum_{k = 1}^{n} sin^{k - 1} x) (\sum_{k = 1}^{n} (2 k - 1) sin^{k - 1} x) cos x d x

,

n \in N

. Then

f_{21} - f_{20}

is equal to ______.

Answer: 41

Solution

Substitute

t = sin x (1 + sin x + sin^{2} x + \dots + sin^{n - 1} x) = \sum_{k = 1}^{n} sin^{\frac{2 k - 1}{2}} x

. Differentiating,

\frac{d t}{d x} = \frac{cos x}{2 sin x} \sum_{k = 1}^{n} (2 k - 1) sin^{k - 1} x

, so

(\sum_{k = 1}^{n} (2 k - 1) sin^{k - 1} x) cos x d x = 2 sin x d t

. Since

\sum_{k = 1}^{n} sin^{k - 1} x = \frac{t}{sin x}

, the integrand becomes

\frac{t}{sin x} \cdot 2 sin x d t = 2 t d t

. As

x

goes from

0

to

\frac{π}{2}

,

t

goes from

0

to

n

, hence

f_{n} = \int_{0}^{n} 2 t d t = n^{2}

. Therefore

f_{21} - f_{20} = 2 1^{2} - 2 0^{2} = 441 - 400 = 41

.

JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Integer answer

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