If int f(x) dx = varphi(x),then int x^5 f(x^3 | vedclass

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(C) Given $\int f(x) dx = \varphi(x)$.Let $I = \int x^5 f(x^3) dx$.Substitute $t = x^3$,then $dt = 3x^2 dx$,which implies $x^2 dx = \frac{1}{3} dt$.Al

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

Vedclass · Answer

(C) Given $\int f(x) dx = \varphi(x)$.Let $I = \int x^5 f(x^3) dx$.Substitute $t = x^3$,then $dt = 3x^2 dx$,which implies $x^2 dx = \frac{1}{3} dt$.Also,$x^5 dx = x^3 \cdot x^2 dx = t \cdot \frac{1}{3} dt$.So,$I = \int t f(t) \cdot \frac{1}{3} dt = \frac{1}{3} \int t f(t) dt$.Using integration by parts,$\int u dv = uv - \int v du$,where $u = t$ and $dv = f(t) dt$ (so $du = dt$ and $v = \varphi(t)$).$I = \frac{1}{3} [t \varphi(t) - \int \varphi(t) dt] + c$.Substituting $t = x^3$ back into the expression:$I = \frac{1}{3} [x^3 \varphi(x^3) - \int \varphi(x^3) \cdot (3x^2 dx)] + c$.$I = \frac{1}{3} x^3 \varphi(x^3) - \int x^2 \varphi(x^3) dx + c$.

If $\int f(x) dx = \varphi(x)$,then $\int x^5 f(x^3) dx = $

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