What is the integral of 1/(3-5sin(x)) ?

The integral of 1/(3-5sin(x)) is 2(-1/8 ln|3tan(x/2)-1|+1/8 ln|tan(x/2)-3|)+C

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integral of 1/(3-5sin(x))

\int \frac{1}{3 - 5 sin ( x )} d x

2 (- \frac{1}{8} ln 3 tan (\frac{x}{2}) - 1 + \frac{1}{8} ln tan (\frac{x}{2}) - 3) + C

Solution steps

\int \frac{1}{3 - 5 sin ( x )} d x

Apply u-substitution

: \int \frac{2}{3 ( 1 + u ^{2} ) - 10 u} d u

= \int \frac{2}{3 ( 1 + u ^{2} ) - 10 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}{3 ( 1 + u ^{2} ) - 10 u} d u

Take the partial fraction of

\frac{1}{3 ( 1 + u ^{2} ) - 10 u} : - \frac{3}{8 ( 3 u - 1 )} + \frac{1}{8 ( u - 3 )}

= 2 \cdot \int - \frac{3}{8 ( 3 u - 1 )} + \frac{1}{8 ( u - 3 )} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (- \int \frac{3}{8 ( 3 u - 1 )} d u + \int \frac{1}{8 ( u - 3 )} d u)

\int \frac{3}{8 ( 3 u - 1 )} d u = \frac{1}{8} ln ∣ 3 u - 1 ∣

\int \frac{1}{8 ( u - 3 )} d u = \frac{1}{8} ln ∣ u - 3 ∣

= 2 (- \frac{1}{8} ln ∣ 3 u - 1 ∣ + \frac{1}{8} ln ∣ u - 3 ∣)

Substitute back

u = tan (\frac{x}{2})

= 2 (- \frac{1}{8} ln 3 tan (\frac{x}{2}) - 1 + \frac{1}{8} ln tan (\frac{x}{2}) - 3)

Add a constant to the solution

= 2 (- \frac{1}{8} ln 3 tan (\frac{x}{2}) - 1 + \frac{1}{8} ln tan (\frac{x}{2}) - 3) + C

(1 + x)^{- 1}

\int \frac{2 x}{x ^{2} - 16} + \frac{1}{x} + 2 d x

y^{^{'}} = x

f (x) = (- 2 x^{2} + 4)^{4}

y = (\frac{( x + 1 )}{( x + 3 ) ^{2}})

What is the integral of 1/(3-5sin(x)) ?
The integral of 1/(3-5sin(x)) is 2(-1/8 ln|3tan(x/2)-1|+1/8 ln|tan(x/2)-3|)+C