evaluate the improper integral sin(x)/sqrt(cos(x)) dx from pi/3 to pi/2?

Question

Help, please.

Answer 1

∫sin(x)/sqrt(cos(x)) dx
= ∫-1/sqrt(cos(x)) dcos(x)
= -2sqrtcos(x), x from pi/3 to pi/2
= √2

Answer 2

Let u = cos x. Then du = -sin x dx
So we have - integral u^-.5du = -2u^.5
So integral is -2(cos x)^.5
= -2(cos90)^.5 + 2(cos 60)^.5
= -2(0)^.5+ 2(1/2)^.5
= sqrt(2)

Answer 3

This is about integration by substitution

Now the differential part of sin(x) = -cos(x)

So if we let u=cos(x) then du/dx = -sin(x)

=>∫(sin(x) / √cos(x)). dx

= ∫(-du/dx / √u).dx

= ∫(-1/√u)du

= ∫-u^(-1/2)du

This integrates using the rule that x^n integrates to [x^(n+1)]/(n+1)

It integrates to [-u^(1/2)]/[1/2] = -2.√u = -2√cos(x)

cos(π/3) =0.5
cos(π/2)=0

So -2√cos(x) from π/3 to π/2

= -2[√cos(π/2) - √cos(π/3)]
= -2[0 - √0.5]
= 2√0.5 = 2/√2
=√2

Answer 4

Let m = cos(x), so dm = sin(x).

Now the integral is m^-(1/2) dm, which integrates as 2 sqrt(m) = 2sqrt(cos(x)).

Now plug in the limits and obtain the final answer.