What is the integral of (sec^2(sqrt(y)))/(2sqrt(y)) ?

The integral of (sec^2(sqrt(y)))/(2sqrt(y)) is tan(sqrt(y))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (sec^2(sqrt(y)))/(2sqrt(y))

\int \frac{sec ^{2} ( y )}{2 y} d y

tan (y) + C

Solution steps

\int \frac{sec ^{2} ( y )}{2 y} d y

Take the constant out:

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

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

Rewrite using trig identities

= \frac{1}{2} \cdot \int \frac{1 + tan ^{2} ( y )}{y} d y

Expand

\frac{1 + tan ^{2} ( y )}{y} : \frac{1}{y} + \frac{tan ^{2} ( y )}{y}

= \frac{1}{2} \cdot \int \frac{1}{y} + \frac{tan ^{2} ( y )}{y} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int \frac{1}{y} d y + \int \frac{tan ^{2} ( y )}{y} d y)

\int \frac{1}{y} d y = 2 y

\int \frac{tan ^{2} ( y )}{y} d y = - 2 y + 2 tan (y)

= \frac{1}{2} (2 y - 2 y + 2 tan (y))

Simplify

\frac{1}{2} (2 y - 2 y + 2 tan (y)) : tan (y)

= tan (y)

Add a constant to the solution

= tan (y) + C

y^{^{'''}} + 3 y^{^{''}} - 4 y = e^{x}

\int \frac{3 y}{x}

\frac{d}{d x} (cos (x) - 1 + \frac{1}{2} x^{2})

f (x) = \frac{1}{4 x ^{2} + 2 x}

(x^{2} + 2) (x - 5)

What is the integral of (sec^2(sqrt(y)))/(2sqrt(y)) ?
The integral of (sec^2(sqrt(y)))/(2sqrt(y)) is tan(sqrt(y))+C