What is the integral of (arccos(x))^2 ?

The integral of (arccos(x))^2 is xarccos^2(x)-2(arccos(x)sqrt(1-x^2)+x)+C

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integral of (arccos(x))^2

\int (arccos (x))^{2} d x

x arccos^{2} (x) - 2 (arccos (x) 1 - x^{2} + x) + C

Solution steps

\int (arccos (x))^{2} d x

Simplify

= \int arccos^{2} (x) d x

Apply Integration By Parts

: x arccos^{2} (x) - \int - \frac{2 x arccos ( x )}{1 - x ^{2}} d x

= x arccos^{2} (x) - \int - \frac{2 x arccos ( x )}{1 - x ^{2}} d x

\int - \frac{2 x arccos ( x )}{1 - x ^{2}} d x = 2 (arccos (x) 1 - x^{2} + x)

= x arccos^{2} (x) - 2 (arccos (x) 1 - x^{2} + x)

Add a constant to the solution

= x arccos^{2} (x) - 2 (arccos (x) 1 - x^{2} + x) + C

y^{^{'}} + 6 y = e^{x}

y = x^{2} 16 - x^{2}

\int_{0}^{\frac{π}{6}} 6 sec^{3} (x) d x

sin (6 (t - \frac{π}{6}))

y = \frac{t}{( t - 4 ) ^{2}}

What is the integral of (arccos(x))^2 ?
The integral of (arccos(x))^2 is xarccos^2(x)-2(arccos(x)sqrt(1-x^2)+x)+C