What is the integral of 1/((cos(2t)+sin(2t))^2) ?

The integral of 1/((cos(2t)+sin(2t))^2) is -1/(2(tan(2t)+1))+C

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integral of 1/((cos(2t)+sin(2t))^2)

\int \frac{1}{( cos ( 2 t ) + sin ( 2 t ) ) ^{2}} d t

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

Solution steps

\int \frac{1}{( cos ( 2 t ) + sin ( 2 t ) ) ^{2}} d t

Rewrite using trig identities

= \int \frac{1}{1 + sin ( 4 t )} d t

Apply u-substitution

: \int \frac{1}{4 ( 1 + sin ( u ) )} d u

= \int \frac{1}{4 ( 1 + sin ( u ) )} d u

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}{1 + sin ( u )} d u

Apply u-substitution

: \int \frac{2}{1 + v ^{2} + 2 v} d v

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

Take the constant out:

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

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

Complete the square

1 + v^{2} + 2 v : (v + 1)^{2}

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

Apply u-substitution

: \int \frac{1}{w ^{2}} d w

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

Apply the Power Rule

: - \frac{1}{w}

= \frac{1}{4} \cdot 2 (- \frac{1}{w})

Substitute back

= \frac{1}{4} \cdot 2 (- \frac{1}{tan ( \frac{4 t}{2} ) + 1})

Simplify

\frac{1}{4} \cdot 2 (- \frac{1}{tan ( \frac{4 t}{2} ) + 1}) : - \frac{1}{2 ( tan ( 2 t ) + 1 )}

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

Add a constant to the solution

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

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What is the integral of 1/((cos(2t)+sin(2t))^2) ?
The integral of 1/((cos(2t)+sin(2t))^2) is -1/(2(tan(2t)+1))+C