What is the integral of (4xsin(y^2))/(y^3) ?

The integral of (4xsin(y^2))/(y^3) is 2x(-(sin(y^2))/(y^2)+\Ci(y^2))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (4xsin(y^2))/(y^3)

\int \frac{4 x sin ( y ^{2} )}{y ^{3}} d y

2 x (- \frac{sin ( y ^{2} )}{y ^{2}} + Ci (y^{2})) + C

Solution steps

\int \frac{4 x sin ( y ^{2} )}{y ^{3}} d y

Take the constant out:

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

= 4 x \cdot \int \frac{sin ( y ^{2} )}{y ^{3}} d y

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{sin ( y ^{2} )}{y ^{3}} = (sin (y^{2})) y^{- 3}

= 4 x \cdot \int sin (y^{2}) y^{- 3} d y

Apply u-substitution

: \int \frac{sin ( u )}{2 u ^{2}} d u

= 4 x \cdot \int \frac{sin ( u )}{2 u ^{2}} d u

Take the constant out:

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

= 4 x \frac{1}{2} \cdot \int \frac{sin ( u )}{u ^{2}} d u

Apply Integration By Parts

: - \frac{sin ( u )}{u} - \int - \frac{cos ( u )}{u} d u

= 4 x \frac{1}{2} (- \frac{sin ( u )}{u} - \int - \frac{cos ( u )}{u} d u)

\int - \frac{cos ( u )}{u} d u = - Ci (u)

= 4 x \frac{1}{2} (- \frac{sin ( u )}{u} - (- Ci (u)))

Substitute back

u = y^{2}

= 4 x \frac{1}{2} (- \frac{sin ( y ^{2} )}{y ^{2}} - (- Ci (y^{2})))

Simplify

4 x \frac{1}{2} (- \frac{sin ( y ^{2} )}{y ^{2}} - (- Ci (y^{2}))) : 2 x (- \frac{sin ( y ^{2} )}{y ^{2}} + Ci (y^{2}))

= 2 x (- \frac{sin ( y ^{2} )}{y ^{2}} + Ci (y^{2}))

Add a constant to the solution

= 2 x (- \frac{sin ( y ^{2} )}{y ^{2}} + Ci (y^{2})) + C

x \to - 1 lim (x + 1)

x \to - 3 lim (\frac{2}{( x + 3 ) ^{6}})

y^{^{'}} = (x + y - 3)^{2}, y (0) = 0

\frac{d}{d x} (8 (e^{x} cos (x) - e^{x} sin (x)))

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

What is the integral of (4xsin(y^2))/(y^3) ?
The integral of (4xsin(y^2))/(y^3) is 2x(-(sin(y^2))/(y^2)+\Ci(y^2))+C