What is the integral of arccos(8x) ?

The integral of arccos(8x) is xarccos(8x)-1/8 sqrt(1-64x^2)+C

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integral of arccos(8x)

\int arccos (8 x) d x

x arccos (8 x) - \frac{1}{8} 1 - 64 x^{2} + C

Solution steps

\int arccos (8 x) d x

Apply Integration By Parts

: x arccos (8 x) - \int - \frac{8 x}{1 - 64 x ^{2}} d x

= x arccos (8 x) - \int - \frac{8 x}{1 - 64 x ^{2}} d x

\int - \frac{8 x}{1 - 64 x ^{2}} d x = \frac{1}{8} 1 - 64 x^{2}

= x arccos (8 x) - \frac{1}{8} 1 - 64 x^{2}

Add a constant to the solution

= x arccos (8 x) - \frac{1}{8} 1 - 64 x^{2} + C

\frac{d}{d x} (12 x^{3} - 6 x^{2})

\frac{y}{x ^{5}}

\int 5 4 x d x

4^{2}

x \to - 2 lim (- x)

What is the integral of arccos(8x) ?
The integral of arccos(8x) is xarccos(8x)-1/8 sqrt(1-64x^2)+C