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We can extend the technique of trigonometric substitution to the hyperbolic functions. Use Theorem 2.20 and the identity cosh2x-sinh2x=1to solve each integral in Exercises 87– 90 with an appropriate hyperbolic substitution x=asinhuoracoshu. (These exercises involve hyperbolic functions.)

x3x2-43/2dx

Short Answer

Expert verified

The value of the integral isx2-8x2-4+C.

Step by step solution

01

Step 1. Given information.

The integral is x3x2-43/2dx.

02

Step 2. Substitution.

Let u=x2

Therefore,

du=2xdx

03

Step 3. Value of the integral.

The value of the integral is ,

x3x2-432dx=u2(u-4)32du=12u(u-4)32du=122v2v2+4dvv=u-4,dv=12u-4du=1dv+4v2dv=v-4v=u-4-4u-4=x2-4-4x2-4=x2-8x2-4+C

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