Question

What is the definite integral of y=sin2x/square root(1+sin^4 x)?the upper limit of integration is pi/2 and the lower limit is 0

Accepted Answer

We have to determine the definite integral of y = sin 2x /sqrt(1 + (sin x)^4), x = 0 to x = pi/2
Int [ sin 2x /sqrt(1 + (sin x)^4) dx]
let 1 + (sin x)^2 = y
dy/dx = 2*sin x* cos x = sin 2x
=> dy = sin 2x dx
Int [ sin 2x /sqrt(1 + (sin x)^4) dx]
=> Int [ 1/sqrt ( 1 + y^2) dy]
=> arcsinh(y)
substitute y = 1 + (sin x)^2
=> arcsinh(1 + (sin x)^2) + C
Between the limits x = 0 and x = pi/2, the definite integral is
arcsinh(1 + 1) - arcsinh(1)
=> arcsinh(2) - arcsinh(1)
The definite integral is arcsinh(2) - arcsinh(1)

Answer

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).

Calculus

What is the definite integral of y=sin2x/square root(1+sin^4 x)? the upper limit of integration is pi/2 and the lower limit is 0

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