If $\int \mathrm{f}(x) \mathrm{d} x=\psi(x)$, then $\int x^5 \mathrm{f}\left(x^3\right) \mathrm{d} x$ is…
- $\frac{1}{3} x^3 \psi\left(x^3\right)-3 \int x^3 \psi\left(x^3\right) \mathrm{d} x+\mathrm{c}$, (where c is a constant of integration)
- $\frac{1}{3}\left(x^3 \psi\left(x^3\right)-\int x^3 \psi\left(x^3\right) \mathrm{d} x\right)+\mathrm{c}$, (where c is a constant of integration)
- $\frac{1}{3} x^3 \psi\left(x^3\right)-\int x^2 \psi\left(x^3\right) \mathrm{d} x+\mathrm{c}$, (where c is a constant of integration)
- $\frac{1}{3}\left(x^3 \psi\left(x^3\right)-\int x^2 \psi\left(x^3\right) \mathrm{d} x\right)+\mathrm{c}$, (where c is a constant of integration)
Solution
Consider $\begin{aligned} \mathrm{I} & =\int x^5 \mathrm{f}\left(x^3\right) \mathrm{d} x \\ & =\int x^3 \cdot x^2 \mathrm{f}\left(x^3\right) d x \end{aligned}$
Let $x^3=\mathrm{t}$ $\Rightarrow 3 x^2 \mathrm{~d} x=\mathrm{dt}$ $\begin{aligned} \therefore \quad & =\int \frac{t}{3} \mathrm{f}(\mathrm{t}) \mathrm{dt} \\ & =\frac{1}{3} \int \mathrm{t} f(\mathrm{t}) \mathrm{dt} \\ & =\frac{1}{3}\left[\mathrm{t} \int \mathrm{f}(\mathrm{t}) \mathrm{dt}-\int \frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{t}) \cdot \int \mathrm{f}(\mathrm{t}) \mathrm{dt}\right] \\ & =\frac{1}{3}\left[\mathrm{t} \cdot \psi(\mathrm{t})-\int \psi(\mathrm{t}) \mathrm{dt}\right]+\mathrm{c} \\ & =\frac{1}{3}\left(x^3 \psi\left(x^3\right)\right)-\int \psi\left(x^3\right) x^2 \cdot \mathrm{~d} x+\mathrm{c}\end{aligned}$
Asked in: MHT CET 2024 (09 May Shift 1)