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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=asinhuorx=acoshu. (These exercises involve hyperbolic functions.)

∫12x2-3dx

Short Answer

Expert verified

The value of the integral is22cosh-16x3+C.

Step by step solution

01

Step 1. Given information.

The given integral is∫12x2-3dx.

02

Step 2. Substitution.

Now, Let x=62cosh(u)

Then,

x=62cosh(u)dx=62sinh(u)du

Since cosh2(u)-1=sinh2(u),

12x2-3=13cosh2(u)-3=33cosh2(u)-1=33sinh2(u)=33sinh(u)

03

Step 3. Value of the integral.

Now, we can write,

∫12x2-3dx=∫22du=22u=22cosh-16x3+C

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