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JEE Main 2013

JEE Main 2013 Mathematics question on Indefinite Integration

If ∫f(x) dx=ψ(x)\int f(x)\, dx = \psi(x), then ∫x5f(x3) dx\int x^5 f(x^3)\, dx is equal to
  1. A.
    13[x3ψ(x3)−∫x2ψ(x3) dx]+c\frac{1}{3}[x^3 \psi(x^3) - \int x^2 \psi(x^3)\, dx] + c
  2. B.
    13x3ψ(x3)−3∫x3ψ(x3) dx+c\frac{1}{3} x^3 \psi(x^3) - 3\int x^3 \psi(x^3)\, dx + c
  3. C.
    13[x3ψ(x3)−∫x2ψ(x3) dx]+c\frac{1}{3}[x^3 \psi(x^3) - \int x^2 \psi(x^3)\, dx] + c
  4. D.
    13[x3ψ(x3)−∫x3ψ(x3) dx]+c\frac{1}{3}[x^3 \psi(x^3) - \int x^3 \psi(x^3)\, dx] + c
Show answer and solution

Correct option: C

Solution

Let t = x³, so dt = 3x² dx. Then:

∫x⁵f(x³)dx = ∫x³·f(x³)·x²dx = (1/3)∫t·f(t)dt

Now integrate ∫t·f(t)dt by parts, using u = t, dv = f(t)dt:
∫t·f(t)dt = t·ψ(t) - ∫ψ(t)dt

Substituting back t = x³:
= x³·ψ(x³) - ∫ψ(x³)dt

But dt = 3x²dx, so ∫ψ(x³)dt = ∫ψ(x³)·3x²dx = 3∫x²ψ(x³)dx

Therefore:
∫x⁵f(x³)dx = (1/3)[x³ψ(x³) - 3∫x²ψ(x³)dx] + c = (1/3)[x³ψ(x³) - ∫x²ψ(x³)dx] + c...

Wait — (1/3)·(−3∫x²ψ(x³)dx) = −∫x²ψ(x³)dx, so:

∫x⁵f(x³)dx = (1/3)x³ψ(x³) − ∫x²ψ(x³)dx + c

This matches option 3: (1/3)[x³ψ(x³) −

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