What is the integral of y/3 arcsin(y^2) ?

The integral of y/3 arcsin(y^2) is 1/6 (y^2arcsin(y^2)+sqrt(1-y^4))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of y/3 arcsin(y^2)

\int \frac{y}{3} arcsin (y^{2}) d y

\frac{1}{6} (y^{2} arcsin (y^{2}) + 1 - y^{4}) + C

Solution steps

\int \frac{y}{3} arcsin (y^{2}) d y

Take the constant out:

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

= \frac{1}{3} \cdot \int y arcsin (y^{2}) d y

Apply Integration By Parts

: \frac{1}{2} y^{2} arcsin (y^{2}) - \int \frac{y ^{3}}{1 - y ^{4}} d y

= \frac{1}{3} (\frac{1}{2} y^{2} arcsin (y^{2}) - \int \frac{y ^{3}}{1 - y ^{4}} d y)

\int \frac{y ^{3}}{1 - y ^{4}} d y = - \frac{1}{2} 1 - y^{4}

= \frac{1}{3} (\frac{1}{2} y^{2} arcsin (y^{2}) - (- \frac{1}{2} 1 - y^{4}))

Simplify

\frac{1}{3} (\frac{1}{2} y^{2} arcsin (y^{2}) - (- \frac{1}{2} 1 - y^{4})) : \frac{1}{6} (y^{2} arcsin (y^{2}) + 1 - y^{4})

= \frac{1}{6} (y^{2} arcsin (y^{2}) + 1 - y^{4})

Add a constant to the solution

= \frac{1}{6} (y^{2} arcsin (y^{2}) + 1 - y^{4}) + C

\int \frac{2 ^{x} 3 ^{x + 1}}{5 ^{x + 2}} d x

\int \frac{1}{e x} d x

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

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

\int \frac{1 + x}{3 x} d x

What is the integral of y/3 arcsin(y^2) ?
The integral of y/3 arcsin(y^2) is 1/6 (y^2arcsin(y^2)+sqrt(1-y^4))+C