What is the integral of (sec(y)tan(y))/(5+4sec^2(y)) ?

The integral of (sec(y)tan(y))/(5+4sec^2(y)) is 1/(2sqrt(5))arctan(2/(sqrt(5))sec(y))+C

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integral of (sec(y)tan(y))/(5+4sec^2(y))

\int \frac{sec ( y ) tan ( y )}{5 + 4 sec ^{2} ( y )} d y

\frac{1}{2 5} arctan (\frac{2}{5} sec (y)) + C

Solution steps

\int \frac{sec ( y ) tan ( y )}{5 + 4 sec ^{2} ( y )} d y

Apply u-substitution

: \int \frac{1}{5 + 4 u ^{2}} d u

= \int \frac{1}{5 + 4 u ^{2}} d u

Apply Integral Substitution

= \int \frac{1}{2 5 ( 1 + v ^{2} )} d v

Take the constant out:

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

= \frac{1}{2 5} \cdot \int \frac{1}{1 + v ^{2}} d v

Use the common integral:

\int \frac{1}{1 + v ^{2}} d v = arctan (v)

= \frac{1}{2 5} arctan (v)

Substitute back

= \frac{1}{2 5} arctan (\frac{2}{5} sec (y))

Add a constant to the solution

= \frac{1}{2 5} arctan (\frac{2}{5} sec (y)) + C

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What is the integral of (sec(y)tan(y))/(5+4sec^2(y)) ?
The integral of (sec(y)tan(y))/(5+4sec^2(y)) is 1/(2sqrt(5))arctan(2/(sqrt(5))sec(y))+C