What is the integral of 1/(sin(x)+cos(x)+3) ?

The integral of 1/(sin(x)+cos(x)+3) is 2/(sqrt(7))arctan(1/(sqrt(7))(2tan(x/2)+1))+C

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integral of 1/(sin(x)+cos(x)+3)

\int \frac{1}{sin ( x ) + cos ( x ) + 3} d x

\frac{2}{7} arctan (\frac{1}{7} (2 tan (\frac{x}{2}) + 1)) + C

Solution steps

\int \frac{1}{sin ( x ) + cos ( x ) + 3} d x

Apply u-substitution

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

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

Complete the square

u^{2} + u + 2 : (u + \frac{1}{2})^{2} + \frac{7}{4}

= \int \frac{1}{( u + \frac{1}{2} ) ^{2} + \frac{7}{4}} d u

Apply u-substitution

: \int \frac{4}{4 v ^{2} + 7} d v

= \int \frac{4}{4 v ^{2} + 7} d v

Take the constant out:

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

= 4 \cdot \int \frac{1}{4 v ^{2} + 7} d v

Apply Integral Substitution

= 4 \cdot \int \frac{1}{2 7 ( w ^{2} + 1 )} d w

Take the constant out:

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

= 4 \cdot \frac{1}{2 7} \cdot \int \frac{1}{w ^{2} + 1} d w

Use the common integral:

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

= 4 \cdot \frac{1}{2 7} arctan (w)

Substitute back

= 4 \cdot \frac{1}{2 7} arctan (\frac{2}{7} (tan (\frac{x}{2}) + \frac{1}{2}))

Simplify

4 \cdot \frac{1}{2 7} arctan (\frac{2}{7} (tan (\frac{x}{2}) + \frac{1}{2})) : \frac{2}{7} arctan (\frac{1}{7} (2 tan (\frac{x}{2}) + 1))

= \frac{2}{7} arctan (\frac{1}{7} (2 tan (\frac{x}{2}) + 1))

Add a constant to the solution

= \frac{2}{7} arctan (\frac{1}{7} (2 tan (\frac{x}{2}) + 1)) + C

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What is the integral of 1/(sin(x)+cos(x)+3) ?
The integral of 1/(sin(x)+cos(x)+3) is 2/(sqrt(7))arctan(1/(sqrt(7))(2tan(x/2)+1))+C