What is the integral of (sec^2(t))/(1+4tan^2(t)) ?

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

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integral of (sec^2(t))/(1+4tan^2(t))

\int \frac{sec ^{2} ( t )}{1 + 4 tan ^{2} ( t )} d t

\frac{1}{2} arctan (2 tan (t)) + C

Solution steps

\int \frac{sec ^{2} ( t )}{1 + 4 tan ^{2} ( t )} d t

Apply u-substitution

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

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

Apply Integral Substitution

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

Substitute back

= \frac{1}{2} arctan (2 tan (t))

Add a constant to the solution

= \frac{1}{2} arctan (2 tan (t)) + C

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