What is the integral of (sec(x)tan(x))/(16+sec^2(x)) ?

The integral of (sec(x)tan(x))/(16+sec^2(x)) is 1/4 arctan((sec(x))/4)+C

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integral of (sec(x)tan(x))/(16+sec^2(x))

\int \frac{sec ( x ) tan ( x )}{16 + sec ^{2} ( x )} d x

\frac{1}{4} arctan (\frac{sec ( x )}{4}) + C

Solution steps

\int \frac{sec ( x ) tan ( x )}{16 + sec ^{2} ( x )} d x

Apply u-substitution

: \int \frac{1}{16 + u ^{2}} d u

= \int \frac{1}{16 + u ^{2}} d u

Apply Integral Substitution

= \int \frac{1}{4 ( v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= \frac{1}{4} arctan (v)

Substitute back

= \frac{1}{4} arctan (\frac{sec ( x )}{4})

Add a constant to the solution

= \frac{1}{4} arctan (\frac{sec ( x )}{4}) + C

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

\int_{- 10}^{10} 100 - x^{2} d x

cos (y) y^{^{'}} + \frac{1}{x} sin (y) = 1

y = 2 x, y = 2, x = 0

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What is the integral of (sec(x)tan(x))/(16+sec^2(x)) ?
The integral of (sec(x)tan(x))/(16+sec^2(x)) is 1/4 arctan((sec(x))/4)+C