• Revision:Further Integration

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Hyperbolic Standard Integrals

 \displaystyle \int \mathrm{sinh}x \hspace5 dx = \mathrm{cosh}x + C

 \displaystyle \int \mathrm{cosh}x \hspace5 dx = \mathrm{sinh}x + C


 \displaystyle \int \mathrm{sech}^2x \hspace5 dx = \mathrm{tanh}x + C


 \displaystyle \int \mathrm{cosech}^2x = -\mathrm{coth}x + C


 \displaystyle \int \mathrm{cosech}x\mathrm{coth}x = -\mathrm{cosech}x + C


 \displaystyle \int \{sech}x = \ln{\mathrm{sech}x \mathrm{tanh}x} + C


 \displaystyle \int \mathrm{cosech}x = \ln{\mathrm{tanh} \frac{x}{2}} + C

 \displaystyle \int \mathrm{tanh}x \hspace5 dx = \ln{\mathrm{sech}x} + C


 \displaystyle \int \mathrm{coth}x \hspace5 dx = \ln{\mathrm{sinh}x} + C

Inverse Trig and Hyperbolic Integrals

 \displaystyle \int \frac{1}{\sqrt{x^2+a^2}} = \mathrm{arsinh}\frac{x}{a} + C

 \displaystyle \int \frac{1}{\sqrt{x^2-a^2}} = \mathrm{arcosh}\frac{x}{a} + C

 \displaystyle \int \frac{1}{1-x^2} = \mathrm{artanh}x + C

 \displaystyle \int \frac{1}{\sqrt{1-x^2}} = \mathrm{arcsin}x + C

 \displaystyle \int \frac{-1}{\sqrt{1-x^2}} = \mathrm{arccos}x + C

 \displaystyle \int \frac{1}{a^2 +x^2} = \frac{1}{a}\mathrm{arctan}\frac{x}{a} + C

 \displaystyle \int \frac{1}{x^2-a^2} = \frac{1}{2a} \ln{|\frac{x-a}{x+a}|} + C

 \displaystyle \int \frac{1}{a^2-x^2}= \frac{1}{2a} \ln{|\frac{a+x}{a-x}|} + C

Reduction Formula

The Reduction Formula is used when integrals can be reduced by generalising the integral for a function to the powers of n. From this you can substitute a value for n and evaluate a definite or indefinite integral relatively easy. They are especially useful when integrating large powers of trigonometric functions.

For example :

Find  \displaystyle \int x^3e^x \hspace5 dx

take  I_n = \displaystyle \int x^ne^x \hspace5 dx

The  I_n refers to the integral where there is a power of n involved.

Evaluating this by integration by parts:

 u = x^n

 \frac{du}{dx} = nx^{n-1}

 \frac{dv}{dx} = e^x

 v = e^x

So

\displaystyle I_n = x^ne^x - \int nx^{n-1}e^x

 \displaystyle I_n = x^ne^x - n \int x^{n-1}e^x

Notice that  \displaystyle \int x^{n-1}e^x = I_{n-1}

becomes  I_n = x^ne^x - nI_{n-1}

From this, we can find  \displaystyle \int x^3e^x \hspace5 dx

by saying "let n = 3".

Therefore,

 I_3 = x^3e^x - 3I_2

 I_2 = x^2e^x - 2I_1

 I_1 = \displaystyle \int xe^x

 u = x

 \frac{du}{dx} = 1

 \frac{dv}{dx} = e^x

  v = e^x

 I_1 = xe^x - \displaystyle \int e^x

 I_1 = xe^x-e^x + C

Therefore,

 I_3 = x^3e^x -3x^2e^x +6(xe^x-e^x) + C

 I_3 = x^3e^x - 3x^2e^x +6xe^x- 6e^x + C

Therefore,

 \displaystyle \int x^3e^x \hspace5 dx = x^3e^x - 3x^2e^x + 9xe^x-9e^x + C

This can be used for  sin^nx and other trigonometric functions.

Length of a Curve

Arc Length AB = \displaystyle \int_{x_a}^{x_b} \left[1+\left(\frac{dy}{dx}\right)^2\right]^{\frac{1}{2}} \hspace5 dx

Arc Length AB = \displaystyle \int_{x_a}^{x_b} \left[1+\left(\frac{dx}{dy}\right)^2\right]^{\frac{1}{2}} \hspace5 dx


Arc Length AB = \displaystyle \int_{x_a}^{x_b} \left[\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2\right]^{\frac{1}{2}} \hspace5 dt

Area of Surface of Revolution

When rotated around the x axis:

Surface Area = \displaystyle \int_{x_a}^{x_b} \displaystyle 2\pi y \left[1+\left(\frac{dy}{dx}\right)^2\right]^{\frac{1}{2}} \hspace5 dx

When using parametric coordinates (for rotation around the x axis):

Surface Area = \displaystyle \int_{x_a}^{x_b} 2\pi y \left[\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2\right]^{\frac{1}{2}} \hspace5 dt

When rotated around the y axis:

Surface Area = \displaystyle \int_{x_a}^{x_b} \displaystyle 2\pi x \left[1+\left(\frac{dy}{dx}\right)^2\right]^{\frac{1}{2}} \hspace5 dx

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