Originally posted by ToO LaZy ^*
evaluate:
1) root(x) / root(1-x) [let u=2x+3]
I'm not seeing the reason for the substitution u = 2x + 3, but the substitution u<sup>2</sup> = 1 - x works well...
Let I = int sqrt(x) / sqrt(1 - x) dx
Now, apply the reverse chain rule, and let u<sup>2</sup> = 1 - x. We see that 2u du = - dx, and x = 1 - u<sup>2</sup>
So, I = int [sqrt(1 - u<sup>2</sup>) / sqrt(u<sup>2</sup>] * (-2u) du = -2 * int u * sqrt(1 - u<sup>2</sup>) / u du
= -2 * int sqrt(1 - u<sup>2</sup> du
Now, again apply the reverse chain rule, this time with u = sin @, and so du = cos @ d@.
Hence, I = -2 * int sqrt(1 - sin<sup>2</sup>@) * cos @ d@ = -2 * int cos<sup>2</sup>@ d@
= -2 * (1 / 2) int 1 + cos2@ d@ = -[@ + (1 /2) * sin 2@] + C = -sin<sup>-1</sup>u + u * sqrt(1 - u<sup>2</sup>) + C
= -sin<sup>-1</sup>[sqrt(1 - x)] + sqrt(x - x<sup>2</sup>) + C
This is a mess ... I think there should be an easier way...
This one is straight forward, once you realise it calls for the substitution x = tan @. The integral becomes
int 1 / sec @ d@ = sin @ + C = x / sqrt(1 + x<sup>2</sup>) + C