integration (1 Viewer)

[azureus88]

need help with this question:

int[1/(x^2(sqrt(x-1))]

how are you meant to do these questions, lik wat to substitute? where do i look for clues?

thanks

 

[Trebla]

Let x = sec²u (this substitution should help remove the square root)
dx = 2sec²u.tan u du
∫ dx / [x²√(x -1)] = ∫ 2sec²u.tan u du / [sec⁴u√(sec²u -1)]
but sec²u - tan²u = 1 => sec²u - 1 = tan²u
= ∫ 2 tan u du / [sec²u.tan u)]
= ∫ 2 sec²u du
= 2∫ cos²u du
but cos 2u = 2cos²u - 1 => 2cos²u = 1 + cos 2u
= ∫ (1 + cos 2u) du
= u + sin 2u / 2 + c
= u + cos u.sin u + c

Now to resubstitute back in terms of x:
x = sec²u
= 1 + tan²u
=> tan u = √(x - 1)
u = tan^-1√(x - 1)
From drawing a right-angled triangle we have
sin u = √(x - 1) / √x
cos u = 1 / √x
=> sin u.cos u = √(x - 1) / x

.: ∫ dx / [x²√(x -1)] = tan^-1√(x - 1) + √(x - 1) / x + c

 

[azureus88]

ok, i see. by the way, why does int{0 ->pi/2} (sqrt(1+sin2x)) = 0.5int{0->pi} (sqrt(1+sinx)) = int{0 ->pi/2} (sqrt(1+sinx))

 

[Trebla]

ok, i see. by the way, why does int{0 ->pi/2} (sqrt(1+sin2x)) = 0.5int{0->pi} (sqrt(1+sinx)) = int{0 ->pi} (sqrt(1+sinx))

For ∫₀^π/2 √(1 + sin 2x) dx
Let u = 2x
du = 2 dx
x = 0, u = 0
x = π/2, u = π
∫₀^π/2 √(1 + sin 2x) dx = 0.5∫₀^π √(1 + sin u) du
= 0.5∫₀^π √(1 + sin x) dx
as u is a dummy variable

That's the first equality but your second equality doesn't seem to make sense... which basically says this integral equals half of itself?...

 

[azureus88]

yeh, its a typo. check the edit. its meant to be pi/2 as opposed to pi

 

[Trebla]

Consider ∫_π/2^π √(1 + sin x) dx
Let u = x - π/2
du = dx
x = π, u = π/2
x = π/2, u = 0
∫_π/2^π √(1 + sin x) dx = ∫₀^π/2 √(1 + sin (u + π/2)) du
But using properties of definite integrals recall that
∫₀^a f(x) dx = ∫₀^a f(a - x) dx
.: ∫₀^π/2 √(1 + sin (u + π/2)) du = ∫₀^π/2 √(1 + sin (π/2 - u + π/2)) du
= ∫₀^π/2 √(1 + sin (π - u)) du
= ∫₀^π/2 √(1 + sin u) du
= ∫₀^π/2 √(1 + sin x) dx
as u is a dummy variable
Hence: ∫_π/2^π √(1 + sin x) dx = ∫₀^π/2 √(1 + sin x) dx
Now ∫₀^π √(1 + sin x) dx = ∫₀^π/2 √(1 + sin x) dx + ∫_π/2^π √(1 + sin x) dx
= 2∫₀^π/2 √(1 + sin x) dx
.: 0.5∫₀^π √(1 + sin x) dx = ∫₀^π/2 √(1 + sin x) dx