If \int f \left(\right. x \left.\right) dx = \Psi \left(\right. x \left.\right) , \textrm{ } then \textrm{ } \int x^{5} f \left(\right. x^{3} \left.\right) dx is equal to :

Engineering

Mathematics

Properties of Definite Integral

Question

If $\int ƒ (x) dx = Ψ (x), then \int x^{5} ƒ (x^{3}) dx$ is equal to :

$\frac{1}{3} [x^{3} Ψ (x^{3}) - \int x^{3} Ψ (x^{3}) dx] + C$

$\frac{1}{3} x^{3} Ψ (x^{3}) - \int x^{2} Ψ (x^{3}) dx + C$

$\frac{1}{3} [x^{3} Ψ (x^{3}) - \int x^{2} Ψ (x^{3}) dx] + C$

$\frac{1}{3} x^{3} Ψ (x^{3}) - 3 \int x^{3} Ψ (x^{3}) dx + C$

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$\int f (x) dx = ψ (x)$

$\int x^{5} f (x^{3}) dx$

put x³ = t $\Rightarrow$ 3x²dx = dt

$\frac{1}{3} \int \underset{I}{\underset{⏟}{t}} \underset{II}{\underset{⏟}{f (t)}} dt = \frac{1}{3} [t \int f (t) dt - \int Ψ (t) dt]$ [Using Integrating by parts]

$= \frac{1}{3} [x^{3} Ψ (x^{3}) - \int Ψ (x^{3}) d (x^{3})]$

$= \frac{1}{3} [x^{3} Ψ (x^{3}) - \int 3 x^{2} Ψ (x^{3}) dy]$

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