MATH help INTEGRAL help?

Question

3. integrate 1 / ((x^2)((sqrt(25 + x^2))) dx by letter x = 5tan(t)

4. integrate 1 / (sqrt(8 + 2x - x^2)) dx

5.integrate from x=4 to x=0 1 / (sqrt(324+x^2)) dx

6. integrate from x=5(sqrt(2))/2 to x=0 (sqrt(25-(x^2))) dx

7. integrate 1 / (x^2)(sqrt(144-x^2)) dx

8. integrate (sqrt(16x-x^2)) dx

9. integrate 44 / (sqrt(40 - 6x - x^2)) dx

Answer 1

3. ∫1 / (x^2 √(25+x^2)) dx
Let x = 5 tan θ, dx = 5 sec^2 θ dθ
= ∫1 / (25 tan^2 θ √(25 + 25 tan^2 θ)) (5 sec^2 θ) dθ
= ∫5 sec^2 θ / (25 tan^2 θ √(25 sec^2 θ)) dθ
= ∫5 sec^2 θ / (125 tan^2 θ sec θ) dθ
= ∫sec θ / 25 tan^2 θ dθ
= ∫(1/cos θ) / (25 sin^2 θ / cos^2 θ) dθ
= ∫cos θ / 25 sin^2 θ dθ
Substitute z = sin θ, dz = cos θ dθ
= ∫1 / (25 z^2) dz
= (1/25) (-1/z) + const
= (-1/25) cosec θ + const
= (-1/25) √ (cot^2 θ + 1) + const
= (-1/25) √ (1 + 1/x^2) + const.

4. ∫1/(√(8 + 2x - x^2)) dx
= ∫1/(√(9 - (x-1)^2)) dx
Let x-1 = 3 sin θ, dx = 3 cos θ dθ.
= ∫1/(√(9 - 9 cos^2 θ)) . 3 cos θ dθ
= ∫(3 cos θ) / (3 sin θ) dθ
Let z = sin θ, dz = cos θ dθ.
= ∫1/z dz
= ln |1/sin θ| + const
= ln |3 / (x-1)| + const
= ln 3 - ln |x-1| + const
= - ln |x-1| + const

5. ∫(4 to 0) 1/√(324 + x^2) dx.
324 = 18^2, so let x = 18 tan θ, dx = 18 sec^2 θ dθ.
= ∫(arctan 2/9 to 0) 1/√(324(1 + tan^2 θ)) (18 sec^2 θ) dθ
= ∫(arctan 2/9 to 0) (18 sec^2 θ / 18 sec θ) dθ
= ∫(arctan 2/9 to 0) sec θ dθ
= ln |sec θ + tan θ| [arctan 2/9 to 0]
= ln |sec 0 + tan 0| - ln |sec arctan 2/9 + 2/9|
Now, draw a right-angled triangle with an angle θ and opposite side 2, adjacent 9 (so θ = arctan 2/9). The hypotenuse is √85, so sec θ = √85 / 9.
= ln 1 - ln ((√85 + 2) / 9)
= 0 - (ln (√85 + 2) - 2 ln 3)
= 2 ln 3 - ln (√85 + 2).

6. ∫(5√2/2 to 0) (√(25-x^2)) dx
Let x = 5 sin θ, dx = 5 cos θ dθ.
= ∫(π/4 to 0) √(25 - 25 sin^2 θ) . 5 cos θ dθ
= ∫(π/4 to 0) . 25 cos^2 θ dθ
= ∫(π/4 to 0) . (25/2) (cos 2θ + 1) dθ
= [(25/4) sin 2θ + (25/2) θ] [π/4 to 0]
= (25/4) sin 0 + (25/2).0 - ((25/4) sin π/2 + (25/2) π/4)
= 0 + 0 - 25/4 - 25π/8.
= -25/8 (2 + π).

7. The bracketing's a bit ambiguous on this one, I will assume it means
∫1 / (x^2 √(144-x^2)) dx
Let x = 12 sin θ, dx = 12 cos θ dθ.
= ∫1 / (144 sin^2 θ √(144 - 144 sin^2 θ)) . (12 cos θ) dθ
= ∫12 cos θ / (144 sin^2 θ . 12 cos θ) dθ
= ∫ 1 / (144 sin^2 &theta); dθ
= 1/144 ∫ cosec^2 θ dθ
= 1/144 (- cot θ) + const
Now if sin θ = x/12, cot θ = √(144-x^2) / x (draw a right triangle with appropriate sides).
= -1/144 radic;(144 - x^2) / x + const.

8. ∫√(16x-x^2) dx
= ∫√(64-(x-8)^2) dx
Let x-8 = 8 sin θ, dx = 8 cos θ dθ.
= ∫√(64 - 64 sin^2 θ) (8 cos θ dθ)
= ∫ 64 cos^2 θ dθ
= ∫ 32 (cos 2θ + 1) dθ
= 32 (sin 2θ) / 2 + 32 θ + const
= 16 sin θ cos θ + 32 θ + const
= (1/4) (x-8) √(64 - (x-8)^2) + 32 arcsin (x/8 - 1) + const
= (1/4) (x-8) √(16x - x^2) + 32 arcsin (x/8 - 1) + const.

9. ∫44 / √(40 - 6x - x^2) dx
= ∫44 / √(49 - (x+3)^2) dx
Let x+3 = 7 sin θ, dx = 7 cos θ dθ
= ∫44 / √(49 - 49 sin^2 θ) (7 cos θ) dθ
= ∫44 (7 cos θ) / (7 cos θ) dθ
= 44 θ + const
= 44 arcsin ((x+3)/7) + const.