Find the integral of acrsinhxdx
just like regular trig functions, use integration by parts to get
x arcsinh(x) - √(x^2+1)
ok thankyou! i wasn't sure if i had to use integration by parts or if there was another method. thankyou!!!
To find the integral of acrsinh(x) dx, we can use integration by parts. The formula for integration by parts is:
∫ u * dv = u * v - ∫ v * du
Let's start by expressing acrsinh(x) as a product of two functions, u and dv, and then find the derivatives and integrals separately.
Let u = acr(x) and dv = sinh(x) dx.
To find du, we need to take the derivative of u with respect to x.
du = d(acr(x)) = 0 - a * dln((x + sqrt(x^2 + 1))) = -a / (x*sqrt(x^2 + 1)) dx
To find v, we need to integrate dv with respect to x.
∫ sinh(x) dx = cosh(x)
Now we can apply the integration by parts formula:
∫ acrsinh(x) dx = u * v - ∫ v * du
= acr(x) * cosh(x) - ∫ cosh(x) * (-a / (x*sqrt(x^2 + 1))) dx
Simplifying further:
∫ acrsinh(x) dx = acr(x) * cosh(x) + a∫cosh(x) / (x*sqrt(x^2 + 1)) dx
The new integral on the right side requires a different method to solve it. We can use a trigonometric substitution.
Let x = sinh(theta), then dx = cosh(theta) dtheta, and sqrt(x^2 + 1) = sqrt(sinh^2(theta) + 1) = cosh(theta).
Substituting these values into the integral:
a∫cosh(x) / (x*sqrt(x^2 + 1)) dx = a∫cosh(theta) / (sinh(theta)*cosh(theta)) * cosh(theta) dtheta
= a∫1 / sinh(theta) dtheta
Using the identity 1/sinh(theta) = coth(theta):
a∫1 / sinh(theta) dtheta = a∫coth(theta) dtheta
To integrate coth(theta), we use another trigonometric substitution.
Let y = e^theta, then dy = e^theta dtheta.
Using the hyperbolic identity coth(theta) = (e^theta + e^(-theta)) / (e^theta - e^(-theta)), we have:
a∫coth(theta) dtheta = a∫[(e^theta + e^(-theta)) / (e^theta - e^(-theta))] dtheta
= a∫[(y + 1/y) / (y - 1/y)] (1/y) dy
Simplifying further:
a∫[(y + 1/y) / (y - 1/y)] (1/y) dy = a∫(y^2 + 1) / (y^2 - 1) dy
Now we can integrate this expression:
a∫(y^2 + 1) / (y^2 - 1) dy = a∫(1 + 2/(y^2 - 1)) dy
= a[y + 2∫(1/(y^2 - 1)) dy]
To integrate 1/(y^2 - 1), we can use partial fraction decomposition:
1/(y^2 - 1) = A/(y - 1) + B/(y + 1)
Multiplying through by (y^2 - 1):
1 = A(y + 1) + B(y - 1)
Expanding and collecting like terms:
1 = (A + B) y + (A - B)
Equate the coefficients of y and the constant terms:
A + B = 0 (from the coefficient of y)
A - B = 1 (from the constant term)
Solving these equations, we find A = 1/2 and B = -1/2.
Substituting these values back into the equation:
1/(y^2 - 1) = 1/2 * 1/(y - 1) - 1/2 * 1/(y + 1)
We can now integrate:
a[y + 2∫(1/(y^2 - 1)) dy] = a[y + 2 * (1/2)ln(|y-1|) - 2 * (1/2)ln(|y+1|)] + C
Simplifying further:
a[y + ln(|y-1|) - ln(|y+1|)] + C = a[y + ln(|y-1| / |y+1|)] + C
Undoing the substitution:
a[y + ln(|e^theta - 1| / |e^theta + 1|)] + C
Replacing y with e^theta:
a[e^theta + ln(|e^theta - 1| / |e^theta + 1|)] + C
Changing back to x:
a[e^acrcosh(x) + ln(|e^acrcosh(x) - 1| / |e^acrcosh(x) + 1|)] + C
This is the final result for the integral of acrsinh(x) dx.