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

The integral of (cos(x))/(3cos(x)-5) is -5/6 arctan(2tan(x/2))+2/3 arctan(tan(x/2))+C

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integral of (cos(x))/(3cos(x)-5)

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

- \frac{5}{6} arctan (2 tan (\frac{x}{2})) + \frac{2}{3} arctan (tan (\frac{x}{2})) + C

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Take the partial fraction of

\frac{- u ^{2} + 1}{( 4 u ^{2} + 1 ) ( u ^{2} + 1 )} : \frac{5}{3 ( 4 u ^{2} + 1 )} - \frac{2}{3 ( u ^{2} + 1 )}

= - \int \frac{5}{3 ( 4 u ^{2} + 1 )} - \frac{2}{3 ( u ^{2} + 1 )} 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

= - (\int \frac{5}{3 ( 4 u ^{2} + 1 )} d u - \int \frac{2}{3 ( u ^{2} + 1 )} d u)

\int \frac{5}{3 ( 4 u ^{2} + 1 )} d u = \frac{5}{6} arctan (2 u)

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

= - (\frac{5}{6} arctan (2 u) - \frac{2}{3} arctan (u))

Substitute back

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

= - (\frac{5}{6} arctan (2 tan (\frac{x}{2})) - \frac{2}{3} arctan (tan (\frac{x}{2})))

Simplify

= - \frac{5}{6} arctan (2 tan (\frac{x}{2})) + \frac{2}{3} arctan (tan (\frac{x}{2}))

Add a constant to the solution

= - \frac{5}{6} arctan (2 tan (\frac{x}{2})) + \frac{2}{3} arctan (tan (\frac{x}{2})) + C

t an g e n t 3 sin (x), at x = \frac{π}{4}

y^{^{'}} + 3 x^{7} y^{9} + \frac{y}{x} = 0

x y^{^{'}} - 2 y = - x

d er i v a t i v e \frac{x + 1}{x}

f a c t or 4 - t^{2}

What is the integral of (cos(x))/(3cos(x)-5) ?
The integral of (cos(x))/(3cos(x)-5) is -5/6 arctan(2tan(x/2))+2/3 arctan(tan(x/2))+C