What is the antiderivative of (arccos(x/6))^2?

1 Answer
Aug 6, 2018

I=xcos^-1(x/6)^2-12cos^-1(x/6)sqrt(1-x^2/36)-2x+C, C in RR

Explanation:

I=intcos^(-1)(x/6)^2dx

Let X=x/6

dx=6dX

So:

I=6intcos^(-1)(X)^2dX

Using intregration by parts :
f'(X)=1, f(X)=X,

g(X)=cos^-1(X)^2, g'(X)=(-2cos^-1(X))/sqrt(1-X^2)

So :

I=6Xcos^(-1)(X)^2+12int(Xcos^(-1)(X))/sqrt(1-X^2)dX

We will use another integration by parts,

g(X)=cos^-1(X), g'(X)=-1/sqrt(1-X^2)

f'(X)=X/sqrt(1-X^2), f(X)=-sqrt(1-X^2) (Here's a proof.)

So :

I=6Xcos^(-1)(X)^2+12(-cos^-1(X)sqrt(1-X^2)-int(-sqrt(1-X^2))/(-sqrt(1-X^2))dX)

=6Xcos^(-1)(X)^2-12cos^-1(X)sqrt(1-X^2)-12int1dX

=6Xcos^(-1)(X)^2-12cos^-1(X)sqrt(1-X^2)-12X+C, C in RR

Substitute back, and finally we have :

I=xcos^-1(x/6)^2-12cos^-1(x/6)sqrt(1-x^2/36)-2x+C, C in RR

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