What is the integral of arccos(sqrt((1+x)/2)) ?

The integral of arccos(sqrt((1+x)/2)) is

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integral of arccos(sqrt((1+x)/2))

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

arccos (\frac{1}{2} (1 + x)) + x arccos (\frac{1}{2} (1 + x)) + arcsin (\frac{1}{2} (1 + x)) - \frac{1}{2} sin (2 arcsin (\frac{1}{2} (1 + x))) + C

Solution steps

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

Apply u-substitution

: \int 4 u arccos (u) d u

= \int 4 u arccos (u) d u

Take the constant out:

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

= 4 \cdot \int u arccos (u) d u

Apply Integration By Parts

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

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

\int - \frac{u ^{2}}{2 1 - u ^{2}} d u = - \frac{1}{4} (arcsin (u) - \frac{1}{2} sin (2 arcsin (u)))

= 4 (\frac{1}{2} u^{2} arccos (u) - (- \frac{1}{4} (arcsin (u) - \frac{1}{2} sin (2 arcsin (u)))))

Substitute back

u = \frac{1 + x}{2}

= 4 \frac{1}{2} (\frac{1 + x}{2})^{2} arccos (\frac{1 + x}{2}) - (- \frac{1}{4} (arcsin (\frac{1 + x}{2}) - \frac{1}{2} sin (2 arcsin (\frac{1 + x}{2}))))

Simplify

4 \frac{1}{2} (\frac{1 + x}{2})^{2} arccos (\frac{1 + x}{2}) - (- \frac{1}{4} (arcsin (\frac{1 + x}{2}) - \frac{1}{2} sin (2 arcsin (\frac{1 + x}{2})))) : arccos (\frac{1}{2} (1 + x)) + x arccos (\frac{1}{2} (1 + x)) + arcsin (\frac{1}{2} (1 + x)) - \frac{1}{2} sin (2 arcsin (\frac{1}{2} (1 + x)))

= arccos (\frac{1}{2} (1 + x)) + x arccos (\frac{1}{2} (1 + x)) + arcsin (\frac{1}{2} (1 + x)) - \frac{1}{2} sin (2 arcsin (\frac{1}{2} (1 + x)))

Add a constant to the solution

= arccos (\frac{1}{2} (1 + x)) + x arccos (\frac{1}{2} (1 + x)) + arcsin (\frac{1}{2} (1 + x)) - \frac{1}{2} sin (2 arcsin (\frac{1}{2} (1 + x))) + C

\int \frac{1}{1 + u ^{2}} d u

y = x^{2}, y = x + 6

\frac{d}{d x} (- ln (- x))

f (x) = \frac{e ^{x}}{2 x + 1}

y^{^{''}} + 11 y^{^{'}} + 30 y = 0

What is the integral of arccos(sqrt((1+x)/2)) ?
The integral of arccos(sqrt((1+x)/2)) is