Integrate, using a trigonometric substitution,  `intx^3sqrt(x^2-4)dx`

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embizze | High School Teacher | (Level 1) Educator Emeritus

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Use a trigonometric substitution for `int x^3 sqrt(x^2-4)dx `

Since this has a form of `u^2-a^2 ` we use the substitution `u=a sec theta ` .

Here `u=x, a=2 ` so `u=2 sec theta, du=2sec theta tan theta `

Note that `sqrt(x^2-4)=sqrt((2 sec theta)^2-4)=sqrt(4tan^2 theta)=2tan theta `

Also `x=u=2sec theta ==> x^3=8 sec^3 theta `

So rewrite the integrand as:

`int (8 sec^3 theta)(2tan theta)(2sec theta tan theta) d theta `

Pull out the constant and rearrange as:

`=32 int tan^2 theta sec^4 theta d theta `

`=32 int tan^2 theta(1+tan^2 theta)sec^2 theta d theta `

`=32[int tan^2 theta sec^2 theta d theta + int tan^4 theta sec^2 theta d theta] `

`=32[(tan^3 theta)/3+(tan^5 theta)/5]+C `

`=32/15 [5tan^3 theta + 3 tan^5 theta]+C `

`=32/15 tan^3 theta[5+3tan^2 theta]+C `

Now `sqrt(x^2-4)=2tan theta ==> tan theta = sqrt(x^2-4)/2 `

`=32/15 (x^2-4)^(3/2)/8[5+3(x^2-4)/4]+C `

Multiply inside the brackets by 4 and outside the brackets by 1/4 to get:

`=1/15(x^2-4)^(3/2)[20+3(x^2-4)]+C `

`=1/15(x^2-4)^(3/2)[3x^2+8]+C `

which is the final answer.

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