# 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

### 2 Answers | Add Yours

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).**

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)**