Question

integration of
$( {x}^{3 } - 3x + 1) \div ( \sqrt{1 - {x}^{2} } ) \: dx$
​

Answer 1

Answer:

$\displaystyle \arcsin x+\tfrac13(7-x^2)\sqrt{1-x^2} + C$

Step-by-step explanation:

Put x = sin u.

Then:

• dx = cos u du
• √(1-x²) = cos u
• x³-3x+1 = 1 - (3-x²)x = 1 - (2+cos²u)sin u = 1 - 2 sin u - cos² u sin u

Putting these together, the integral is

$\displaystyle\int\frac{x^3-3x+1}{\sqrt{1-x^2}}\,dx\\\\=\int(1-2\sin u-\cos^2u\,\sin u)\,du\\\\=\int du-2\int\sin u\,du+\int\cos^2u\,d(\cos u)\\\\=u+2\cos u+\tfrac13\cos^3u+C\\\\=u+\tfrac13(6+\cos^2u)\cos u+C\\\\=u+\tfrac13(7-\sin^2u)\cos u+C\\\\=\arcsin x+\tfrac13(7-x^2)\sqrt{1-x^2} + C$