How do you find the integral of 3x (sqrt(81-x^2))3x(√81−x2)?

3 Answers
Jim H · Gió
Apr 20, 2015

Have a look:
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The answer using this method will look like the other, if you use:
cos(arcsin(x/9)) = sqrt(1- x^2/81)cos(arcsin(x9))=√1−x281 So

-729 cos^3(arcsin(x/9)) = -729 (sqrt(1- x^2/81))^3 = -(9sqrt(1- x^2/81))^3 = -(sqrt (81-x^2))^3−729cos3(arcsin(x9))=−729(√1−x281)3=−(9√1−x281)3=−(√81−x2)3

Massimiliano
Apr 20, 2015

The answer is: -sqrt((81-x^2)^3)+c−√(81−x2)3+c

Remembering that:

int[f(x)]^n*f'(x)dx=[f(x)]^(n+1)/(n+1)+c,

than:

int3xsqrt(81-x^2)dx=-3/2int-2x(81-x^2)^(1/2)dx=

=-3/2(81-x^2)^(1/2+1)/(1/2+1)+c=-3/2(81-x^2)^(3/2)/(3/2)+c=

=-sqrt((81-x^2)^3)+c

or, if you want

=-(81-x^2)sqrt(81-x^2)+c.

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