What is the integral of 1/([sin(x)+cos(x)]^2) ?

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

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integral of 1/([sin(x)+cos(x)]^2)

\int \frac{1}{[ sin ( x ) + cos ( x ) ] ^{2}} d x

- \frac{1}{tan ( x ) + 1} + C

Solution steps

\int \frac{1}{[ sin ( x ) + cos ( x ) ] ^{2}} d x

Simplify

\frac{1}{[ sin ( x ) + cos ( x ) ] ^{2}} : \frac{1}{1 + sin ( 2 x )}

= \int \frac{1}{1 + sin ( 2 x )} d x

Apply u-substitution

: \int \frac{1}{2 ( 1 + sin ( u ) )} d u

= \int \frac{1}{2 ( 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}{2} \cdot \int \frac{1}{1 + sin ( u )} d u

Apply u-substitution

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

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{w}

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

Substitute back

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

Simplify

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

= - \frac{1}{tan ( x ) + 1}

Add a constant to the solution

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

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