How do you evaluate by completing the square #int_-1^0dx/sqrt(1-2x-x^2)#?

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Answer 1 · 2017-09-14T13:50:56

# pi/4.#

Here is another way to solve the Problem :

Recall that, #int_a^bf(x)dx=int_a^bf(a+b-x)dx.............(star).#

In #I=int_(-1)^0dx/sqrt(1-2x-x^2),# we have,

#a=-1, b=0, f(x)=1/sqrt(1-2x-x^2).#

#:. f(a+b-x)=f(-1-x)=1/sqrt{1-2(-1-x)-(-1-x)^2},#

#=1/sqrt{1+2+2x-(1+2x+x^2)},#

#=1/sqrt(2-x^2).#

Therefore, by #(star), I=int_(-1)^0{1/sqrt(2-x^2)}dx,#

#=[arc sin(x/sqrt2)]_(-1)^0,#

#=arc sin 0-arc sin(-1/sqrt2),#

#=arc sin(1/sqrt2).#

# rArr I=pi/4.#

Enjoy maths.!