$\int f(x)\mathrm{d}x=g(x)$, then $\int {x}^5f({x}^3)\mathrm{d}x$

Quick Solution

Given:

\( \int f(x)\, dx = g(x) \)

Required: \( \int x^5 f(x^3)\, dx \)

Use substitution:

Let \( u = x^3 \Rightarrow du = 3x^2\, dx \Rightarrow dx = \frac{du}{3x^2} \)

Now rewrite the integral:

\[ \int x^5 f(x^3)\, dx = \int x^5 f(u) \cdot \frac{du}{3x^2} = \frac{1}{3} \int x^3 f(u)\, du \]

But \( x^3 = u \), so:

\[ \frac{1}{3} \int u f(u)\, du \]

Now integrate by parts or use the identity:

\[ \int u f(u)\, du = u g(u) - \int g(u)\, du \]

Final answer:

\[ \int x^5 f(x^3)\, dx = \frac{1}{3} \left[ x^3 g(x^3) - \int g(x^3) \cdot 3x^2\, dx \right] = x^3 g(x^3) - \int x^2 g(x^3)\, dx \]

\[ \boxed{ \int x^5 f(x^3)\, dx = x^3 g(x^3) - \int x^2 g(x^3)\, dx } \]