We'll solve the integral using
substitution.
Let (sin x)^2 =
t.
If we'll differentiate both sides, we'll
get:
2 sin x*cos x dx = dt
But
2 sin x* cos x = sin 2x => sin 2x dx = dt
We'll
re-write the integral in t:
Int sin 2x dx/sqrt[1+(sin x)^4]
= Int dt/sqrt(1 + t^2)
Int dt/sqrt(1 + t^2) = ln [t +
sqrt(1+t^2)]
We'll apply Leibniz Newton
formula.
Since the variable x was changed, we'll change the
value of limits of integration, too.
If x = 0 =>
(sin 0)^2 = 0 = t
If x = pi/2 => (sin pi/2)^2 = 1 =
t
Int dt/sqrt(1 + t^2) = F(1) -
F(0)
F(0) = ln [0 + sqrt(1+0^2)] = ln 1 =
0
F(1) = ln [1 +
sqrt(1+1^2)]
F(1) = ln
(1+sqrt2)
F(1) - F(0) = ln
(1+sqrt2)
The requested definite integral, if
x = 0 to x = pi/2, is: Int sin 2x dx/sqrt[1+(sin x)^4] = ln
(1+sqrt2).
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