How do you find #int1/sqrt(28-12x-x^2)dx#?

Question

1347 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-22T14:42:59

#int1/sqrt(28-12x-x^2)dx=arc sin((x+6)/8)+c#

Here ,

#I=int1/sqrt(28-12x-x^2)dx#

Now,

#color(green)(28-12x-x^2=64-36-12x-x^2=64-(x^2+12x+36)#

#:.I=int1/sqrt(64-(x+6)^2)dx#

Subst. #color(blue)(x+6=8sinu=>dx=8cosudu#

#and color(blue)(sinu=(x+6)/8=>u=arc sin((x+6)/8)#

So ,

#I=int1/sqrt(64-64sin^2u)*8cosudu#

#=int1/(8cosu)xx 8cosudu#

#=int1du#

#=u+c ,where, color(blue)(u=arc sin((x+6)/8)#

#:.I=arc sin((x+6)/8)+c#