What is the integral of arccos(4x) ?

The integral of arccos(4x) is xarccos(4x)-1/4 sqrt(1-16x^2)+C

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integral of arccos(4x)

\int arccos (4 x) d x

x arccos (4 x) - \frac{1}{4} 1 - 16 x^{2} + C

Solution steps

\int arccos (4 x) d x

Apply Integration By Parts

: x arccos (4 x) - \int - \frac{4 x}{1 - 16 x ^{2}} d x

= x arccos (4 x) - \int - \frac{4 x}{1 - 16 x ^{2}} d x

\int - \frac{4 x}{1 - 16 x ^{2}} d x = \frac{1}{4} 1 - 16 x^{2}

= x arccos (4 x) - \frac{1}{4} 1 - 16 x^{2}

Add a constant to the solution

= x arccos (4 x) - \frac{1}{4} 1 - 16 x^{2} + C

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

\frac{\partial}{\partial x} (x^{2} + y^{2} - z)

\int 3 - \frac{1}{3} \cdot (x - 4)^{2} d x

\int \frac{3 x ^{2} + 1}{( x ^{2} + 1 ) ^{3}} d x

\int_{0}^{2} \frac{x ^{2} - 2}{x + 1} d x

What is the integral of arccos(4x) ?
The integral of arccos(4x) is xarccos(4x)-1/4 sqrt(1-16x^2)+C