If $\int f(x) d x=\Psi(x)$, then $\int x^{5} f\left(x^{3}\right) d x$ is equal to

(B) $\frac{1}{3} \mathrm{x}^{3} \Psi\left(\mathrm{x}^{3}\right)-\int \mathrm{x}^{2} \Psi\left(\mathrm{x}^{3}\right) \mathrm{dx}+\mathrm{C}$

(A) $\frac{1}{3} \mathrm{x}^{3} \Psi\left(\mathrm{x}^{3}\right)-3 \int \mathrm{x}^{3} \Psi\left(\mathrm{x}^{3}\right) \mathrm{dx}+\mathrm{C}$

(C) $\frac{1}{3}\left[\mathrm{x}^{3} \Psi\left(\mathrm{x}^{3}\right)-\int \mathrm{x}^{3} \Psi\left(\mathrm{x}^{3}\right) \mathrm{dx}\right]+\mathrm{C}$

(D) $\frac{1}{3}\left[\mathrm{x}^{3} \Psi\left(\mathrm{x}^{3}\right)-\int \mathrm{x}^{2} \Psi\left(\mathrm{x}^{3}\right) \mathrm{dx}\right]+\mathrm{C}$

JEE Main · 2013 · 07 Apr 2013 · Offline · Indefinite Integrals

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\int f (x) d x = Ψ (x)

, then

\int x^{5} f (x^{3}) d x

is equal to

(A) $\frac{1}{3} x^{3} Ψ (x^{3}) - 3 \int x^{3} Ψ (x^{3}) dx + C$
(B) $\frac{1}{3} x^{3} Ψ (x^{3}) - \int x^{2} Ψ (x^{3}) dx + C$ ✓ Correct
(C) $\frac{1}{3} [x^{3} Ψ (x^{3}) - \int x^{3} Ψ (x^{3}) dx] + C$
(D) $\frac{1}{3} [x^{3} Ψ (x^{3}) - \int x^{2} Ψ (x^{3}) dx] + C$

Answer: (B)

Solution

Set

t = x^{3}

, so

d t = 3 x^{2} d x

, and split off the right pieces:

\int x^{5} f (x^{3}) d x = \int x^{3} f (x^{3}) x^{2} d x = \frac{1}{3} \int t f (t) d t

. Integrate by parts with

u = t

and

d v = f (t) d t

, where

\int f (t) d t = Ψ (t)

\int t f (t) d t = t Ψ (t) - \int Ψ (t) d t

. Returning to

x

, the term

\int Ψ (t) d t

becomes

\int Ψ (x^{3}) 3 x^{2} d x

, and the whole result tidies to

\frac{1}{3} x^{3} Ψ (x^{3}) - \int x^{2} Ψ (x^{3}) d x + C

JEE Main · 2013 · 07 Apr 2013 · Offline · Single correct option

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JEE Main · 2013 · 07 Apr 2013 · Offline · Indefinite Integrals

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