Q.251. Indefinite and definite integrals.

Question 251.
1) Find ∫ dx/(2 + √x)^3
2) Evaluate  (r=0 to 2) r √[5 - √(4 - r^2)] dr.

Answer 251.
1)
∫dx/(2 + √x)^3

Let 2 + √x = u
=> √x = u - 2
=> x = u^2 - 4u + 4
=> dx = (2u - 4) du

=> integral
= ∫ (2u - 4) / u^3 du
= 2 ∫ du/u^2 - 4 ∫ du/u^3
= - 2/u + 2/u^2 + c
= 2 [1/(2 + √x)^2 - 1/(2 + √x)] + c
= - 2 (1 + √x) / (2 + √x)^2 + c.

2)
∫ (r=0 to 2) r √[5 - √(4 - r^2)] dr

Let √[5 - √(4 - r^2)] = u
=> 5 - √(4 - r^2) = u^2
=> √(4 - r^2) = 5 - u^2
=> 4 - r^2 = (5 - u^2)^2 = 25 - 10u^2 + u^4
=> - 2r dr = (-20u + 4u^3) du
=> r dr = (10u - 2u^3) du

Also, r = 0 => u = √3
and r = 2 => u = √5

=> integral (u = √3 to √5)
= ∫ (10u - 2u^3) * u du
= 10u^3/3 - (2/5) u^5
= [ (50/3)√5 - 10√5] - [(10/3) (√3)^3 - (2/5) (√3)^5]
= (20/3)√5 - [10√3 - (18/5)√3]
= (20/3)√5 - (32/5)√3.
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