Answer to Question #332673 in Calculus for Olga

Question

Evaluate the following integral:

∫ 3x √ 1 − 2x^2 dx

 

a.    2 √ ( 1 + 2x^2 )^3 + c

b.   √ ( 1 + 2x^2 )^3 + c

c.    1 / 2 √ ( 1 + 2 x^2)^3 + c

d.   1 / 2  3√(1 + 2x^2 ) ^3 + c

Answer 1

$\int 3x\sqrt{1-2x^2}dx = [u=1-2x^2, du = -4xdx \rightarrow xdx=-\cfrac{du}{4}]=-\cfrac{3}{4}\int\sqrt udu = - \cfrac{3\cdot2 u^{\frac{3}{2}}}{4\cdot 3} + C= -\cfrac{(1-2x^2)^{\frac{3}{2}}}{2}+ C$