What is the integral of 1/(5+4sin(x)) ?

The integral of 1/(5+4sin(x)) is 2/3 arctan(5/3 tan(x/2)+4/3)+C

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integral of 1/(5+4sin(x))

\int \frac{1}{5 + 4 sin ( x )} d x

\frac{2}{3} arctan (\frac{5}{3} tan (\frac{x}{2}) + \frac{4}{3}) + C

Solution steps

\int \frac{1}{5 + 4 sin ( x )} d x

Apply u-substitution

: \int \frac{2}{5 ( 1 + u ^{2} ) + 8 u} d u

= \int \frac{2}{5 ( 1 + u ^{2} ) + 8 u} d u

Take the constant out:

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

= 2 \cdot \int \frac{1}{5 ( 1 + u ^{2} ) + 8 u} d u

Expand

5 (1 + u^{2}) : 5 + 5 u^{2}

= 2 \cdot \int \frac{1}{5 + 5 u ^{2} + 8 u} d u

Complete the square

5 + 5 u^{2} + 8 u : 5 (u + \frac{4}{5})^{2} + \frac{9}{5}

= 2 \cdot \int \frac{1}{5 ( u + \frac{4}{5} ) ^{2} + \frac{9}{5}} d u

Apply u-substitution

: \int \frac{5}{25 v ^{2} + 9} d v

= 2 \cdot \int \frac{5}{25 v ^{2} + 9} d v

Take the constant out:

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

= 2 \cdot 5 \cdot \int \frac{1}{25 v ^{2} + 9} d v

Apply Integral Substitution

= 2 \cdot 5 \cdot \int \frac{1}{15 ( w ^{2} + 1 )} d w

Take the constant out:

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

= 2 \cdot 5 \cdot \frac{1}{15} \cdot \int \frac{1}{w ^{2} + 1} d w

Use the common integral:

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

= 2 \cdot 5 \cdot \frac{1}{15} arctan (w)

Substitute back

= 2 \cdot 5 \cdot \frac{1}{15} arctan (\frac{5}{3} (tan (\frac{x}{2}) + \frac{4}{5}))

Simplify

2 \cdot 5 \cdot \frac{1}{15} arctan (\frac{5}{3} (tan (\frac{x}{2}) + \frac{4}{5})) : \frac{2}{3} arctan (\frac{5}{3} tan (\frac{x}{2}) + \frac{4}{3})

= \frac{2}{3} arctan (\frac{5}{3} tan (\frac{x}{2}) + \frac{4}{3})

Add a constant to the solution

= \frac{2}{3} arctan (\frac{5}{3} tan (\frac{x}{2}) + \frac{4}{3}) + C

5^{2 x} x

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\int cos^{2} (θ) d θ

What is the integral of 1/(5+4sin(x)) ?
The integral of 1/(5+4sin(x)) is 2/3 arctan(5/3 tan(x/2)+4/3)+C