What is the integral of-3arcsin(2+z) ?

The integral of-3arcsin(2+z) is -3(zarcsin(2+z)+sqrt(-z^2-4z-3)+2arcsin(z+2))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of-3arcsin(2+z)

\int - 3 arcsin (2 + z) d z

- 3 (z arcsin (2 + z) + - z^{2} - 4 z - 3 + 2 arcsin (z + 2)) + C

Solution steps

\int - 3 arcsin (2 + z) d z

Take the constant out:

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

= - 3 \cdot \int arcsin (2 + z) d z

Apply Integration By Parts

: z arcsin (2 + z) - \int \frac{z}{- z ^{2} - 4 z - 3} d z

= - 3 (z arcsin (2 + z) - \int \frac{z}{- z ^{2} - 4 z - 3} d z)

\int \frac{z}{- z ^{2} - 4 z - 3} d z = - - z^{2} - 4 z - 3 - 2 arcsin (z + 2)

= - 3 (z arcsin (2 + z) - (- - z^{2} - 4 z - 3 - 2 arcsin (z + 2)))

Simplify

= - 3 (z arcsin (2 + z) + - z^{2} - 4 z - 3 + 2 arcsin (z + 2))

Add a constant to the solution

= - 3 (z arcsin (2 + z) + - z^{2} - 4 z - 3 + 2 arcsin (z + 2)) + C

\int \frac{x ^{2}}{4 - 9 x ^{6}} d x

\int \frac{2 x ^{5} + 3 x ^{4} - 5 x ^{2} + 8}{x ^{2}} d x

\int cos (\frac{x}{3}) sin (\frac{x}{3}) d x

\int \frac{tan ^{3} ( x )}{sec ^{4} ( x )} d x

\int (2 x e^{- x^{2}}) e^{x^{2}} d x

What is the integral of-3arcsin(2+z) ?
The integral of-3arcsin(2+z) is -3(zarcsin(2+z)+sqrt(-z^2-4z-3)+2arcsin(z+2))+C