How do you find the antiderivative of #int x^2/sqrt(4-x^2)dx#?

Question

1326 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-22T19:06:09

# 2arc sin(x/2)-x/2sqrt(4-x^2)+C.#

We know that,

# (1) : intsqrt(a^2-x^2)dx=x/2sqrt(a^2-x^2)+a^2/2arc sin(x/a)+c_1.#

# (2) : int1/sqrt(a^2-x^2)dx=arc sin(x/a)+c_2.#

Hence, #I=intx^2/sqrt(4-x^2)dx#

#=-int(-x^2)/sqrt(4-x^2)dx=-int{(4-x^2)-4}/sqrt(4-x^2)dx,#

#=-int(4-x^2)/sqrt(4-x^2)dx+4int1/sqrt(4-x^2)dx,#

#=-intsqrt(4-x^2)dx+4arc sin(x/2),............[because, (2)]#

#=-{x/2sqrt(4-x^2)+4/2arc sin(x/2)}+4arc sin (x/2),#

#=2arc sin(x/2)-x/2sqrt(4-x^2)+C.#

Enjoy Maths.!