Question

Integration of x2sinx3

Answer 1

(x2)(sinx3)dx=−1/3cosx3+C

Explanation:

∫(x2)(sinx3)dx

Let u=x3 so that du=3x2dx

Out integral becomes:

1/3∫(sinx3)3x2dx=1/3∫sinudu

=1/3(−cosu)+C=−1/3cosx3+C

Check the answer by differentiating:

ddx(−1/3cosx3+C)=−1/3−sin(x3)⋅3x2

=sin(x3)⋅x2

pls mark brainliest