Animated Solution for Mathematics - Indefinite Integration: If $\int f(x)dx=\psi (x)$, then $\int x^{5}f(x^3)dx$ is equal to

Welcome, future engineer! Today, we are going to peel back the layers of a problem that, at first glance, looks like a tangled mess of variables. We are asked to evaluate the integral $I=\int x^{5}f(x^3)dx$, given that $\int f(x)dx=\psi (x)$.

When you see a problem like this, it is natural to feel a bit intimidated. You have a function $f$ with an argument $x^3$, and a coefficient $x^5$ sitting outside. The mismatch is the first thing your brain should flag.

But in the world of JEE Advanced, a mismatch is not a roadblock; it is a hint.

The secret to this problem lies in the power of $x$. We have $x^5$ and $x^3$. If we think about the derivative of $x^3$, we get $3x^2$.

Notice that $x^5$ can be written as $x^3\cdot x^2$. By splitting the term, we create the $x^2$ that we desperately need to facilitate a substitution.

So, we rewrite our integral as:

Now, the structure is clear. We have a function of $x^3$ and the derivative of $x^3$ (the $x^2$ term). This is the green light to proceed to substitution.

Let us introduce a new variable, $t$, where $t=x^3$. Differentiating both sides with respect to $x$, we get $dt=3x^{2}dx$, which implies $x^{2}dx=\frac{1}{3}dt$.

Now, watch the magic happen. We substitute these into our integral:

Pulling the constant $\frac{1}{3}$ out, we are left with:

This is a beautiful, clean integral. We have transformed a complex-looking expression into a standard product of an algebraic function $t$ and a function $f(t)$ whose integral we already know.

Now, we apply the Integration by Parts formula: $\int udv=uv-\int vdu$. We choose $u=t$ and $dv=f(t)dt$.

Why? Because $u=t$ becomes $du=dt$ upon differentiation, which simplifies the expression, and we know that $\int f(t)dt=\psi (t)$.

Applying the formula, we get:

We are almost there! The structure is emerging.

Now, we must return to the world of $x$. We substitute $t=x^3$ back into our expression. The term $t\psi (t)$ becomes $x^3\psi (x^3)$.

For the integral part, we have $\int \psi (t)dt$. Since $t=x^3$, we know $dt=3x^{2}dx$. Substituting these back, we get:

The $3$ and the $\frac{1}{3}$ cancel out perfectly, leaving us with the final result:

You see? By breaking the problem down, respecting the substitution, and trusting the Integration by Parts formula, we turned a daunting expression into a simple, elegant solution. Keep practicing this mindset—look for the hidden derivatives, and the path will always reveal itself.