How do you find the integral ∫x3⋅√9−x2dx ?
1 Answer
Aug 29, 2014
=(9−x2)525−3(9−x2)32+c , wherec is a constantExplanation :
=∫x3⋅√9−x2dx Using Integration by Substitution,
let's assume
9−x2=t2 , then
−2xdx=2tdt ⇒xdx=−tdt
=∫−(9−t2)t2dt
=∫(t4−9t2)dt
=∫t4dt−9∫t2dt
=t55−9t33+c , wherec is a constant
=t55−3t3+c , wherec is a constantSubstituting
t back,
=(9−x2)525−3(9−x2)32+c , wherec is a constant