int x^2 sec(x^3) , dx = | vedclass

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(C) Let $I = \int x^2 \sec(x^3) \, dx$. Substitute $t = x^3$. Then,$dt = 3x^2 \, dx$,which implies $x^2 \, dx = \frac{1}{3} \, dt$. Substituting these

Vedclass · Accepted Answer

$\frac{1}{3} \log|\sec(x^3) + 	an(x^3)| + C$

Vedclass · Answer

(C) Let $I = \int x^2 \sec(x^3) \, dx$. Substitute $t = x^3$. Then,$dt = 3x^2 \, dx$,which implies $x^2 \, dx = \frac{1}{3} \, dt$. Substituting these into the integral: $I = \int \sec(t) \cdot \frac{1}{3} \, dt$ $I = \frac{1}{3} \int \sec(t) \, dt$ Using the standard integral formula $\int \sec(t) \, dt = \log|\sec(t) + 	an(t)| + C$: $I = \frac{1}{3} \log|\sec(t) + 	an(t)| + C$ Substituting $t = x^3$ back: $I = \frac{1}{3} \log|\sec(x^3) + 	an(x^3)| + C$. Thus,the correct option is $C$.

$\int x^2 \sec(x^3) \, dx = $

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