What is the integral of 2arccos(4x) ?

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

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integral of 2arccos(4x)

\int 2 arccos (4 x) d x

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

Solution steps

\int 2 arccos (4 x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int arccos (4 x) d x

Apply Integration By Parts

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

= 2 (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}

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

Add a constant to the solution

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

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