Indefinite Integration
Integration by Substitution
Grade 12
Question:
<p>[JEE Main 2021] If \(\displaystyle\int f(x)\,dx=\psi(x)\), then \(\displaystyle\int x^5 f(x^3)\,dx\) equals</p>
<li>\(\dfrac{1}{3}\!\left[x^3\psi(x^3)-3\displaystyle\int x^2\psi(x^3)\,dx\right]+C\)</li>
<li>\(\dfrac{1}{3}\!\left[x^3\psi(x^3)-\displaystyle\int x^2\psi(x^3)\,dx\right]+C\)</li>
<li>\(\dfrac{1}{3}\!\left[x^3\psi(x^3)-3\displaystyle\int x^2\psi(x^3)\,dx\right]+C\)</li>
<li>\(\dfrac13 x^3\psi(x^3)+\dfrac13\displaystyle\int x^2\psi(x^3)\,dx+C\)</li>
Step-by-Step Solution
Key Concept: Write x^5f(x^3)dx = x^3 \cdot [x^2f(x^3)dx]. Note x^2f(x^3)dx = (1/3)d(\psi(x^3)). Then integrate by parts.
<p><strong>Key insight:</strong> Since $\int f(x)\,dx=\psi(x)$, we have $\int f(x^3)\cdot 3x^2\,dx=\psi(x^3)$.</p>
<p>So $x^2 f(x^3)\,dx = \dfrac{1}{3}\,d(\psi(x^3))$.</p>
<p>Write $x^5 f(x^3) = x^3\cdot[x^2 f(x^3)]$. Integrate by parts with $u=x^3$, $dv=x^2f(x^3)\,dx=\frac13 d\psi(x^3)$:</p>
<p>$$\int x^5 f(x^3)\,dx = \frac{x^3}{3}\psi(x^3)-\int\frac{\psi(x^3)}{3}\cdot 3x^2\,dx+C$$</p>
<p>$$= \frac{x^3\psi(x^3)}{3}-\int x^2\psi(x^3)\,dx+C = \frac13\!\left[x^3\psi(x^3)-3\int x^2\psi(x^3)\,dx\right]+C$$</p>
<p>Answer: <strong>(C)</strong></p>
Correct Answer: C