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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 = $

Similar Questions

$\int \frac{\log \left(x+\sqrt{1+x^2}\right)}{\sqrt{1+x^2}} \,dx = \frac{1}{2}(g(x))^2 + C$,(where $C$ is the constant of integration). Then $g(x) =$

$\int \frac{x}{x^4 + x^2 + 1} dx$ is equal to

$\int \frac{\cot x}{\log(\sin x)} \, dx = $

Find the value of $k$ if $\int \cos ^k(x) \sin (x) d x = \frac{-1}{4} \cos ^4(x) + C$.

$\int \frac{\log (\cot x)}{\sin 2 x} \,d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module