What is the integral of (cos(θ))/(5+4cos(θ)) ?

The integral of (cos(θ))/(5+4cos(θ)) is 2(1/4 arctan(tan(θ/2))-5/12 arctan(1/3 tan(θ/2)))+C

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integral of (cos(θ))/(5+4cos(θ))

\int \frac{cos ( θ )}{5 + 4 cos ( θ )} d θ

2 (\frac{1}{4} arctan (tan (\frac{θ}{2})) - \frac{5}{12} arctan (\frac{1}{3} tan (\frac{θ}{2}))) + C

Solution steps

\int \frac{cos ( θ )}{5 + 4 cos ( θ )} d θ

Apply u-substitution

: \int \frac{2 ( 1 - u ^{2} )}{( u ^{2} + 9 ) ( 1 + u ^{2} )} d u

= \int \frac{2 ( 1 - u ^{2} )}{( u ^{2} + 9 ) ( 1 + u ^{2} )} 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 - u ^{2}}{( u ^{2} + 9 ) ( 1 + u ^{2} )} d u

Expand

\frac{1 - u ^{2}}{( u ^{2} + 9 ) ( 1 + u ^{2} )} : \frac{1}{( u ^{2} + 9 ) ( 1 + u ^{2} )} - \frac{u ^{2}}{( u ^{2} + 9 ) ( 1 + u ^{2} )}

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{1}{( u ^{2} + 9 ) ( 1 + u ^{2} )} d u - \int \frac{u ^{2}}{( u ^{2} + 9 ) ( 1 + u ^{2} )} d u)

\int \frac{1}{( u ^{2} + 9 ) ( 1 + u ^{2} )} d u = - \frac{1}{24} arctan (\frac{u}{3}) + \frac{1}{8} arctan (u)

\int \frac{u ^{2}}{( u ^{2} + 9 ) ( 1 + u ^{2} )} d u = \frac{3}{8} arctan (\frac{u}{3}) - \frac{1}{8} arctan (u)

= 2 (- \frac{1}{24} arctan (\frac{u}{3}) + \frac{1}{8} arctan (u) - (\frac{3}{8} arctan (\frac{u}{3}) - \frac{1}{8} arctan (u)))

Substitute back

u = tan (\frac{θ}{2})

= 2 (- \frac{1}{24} arctan (\frac{tan ( \frac{θ}{2} )}{3}) + \frac{1}{8} arctan (tan (\frac{θ}{2})) - (\frac{3}{8} arctan (\frac{tan ( \frac{θ}{2} )}{3}) - \frac{1}{8} arctan (tan (\frac{θ}{2}))))

Simplify

2 (- \frac{1}{24} arctan (\frac{tan ( \frac{θ}{2} )}{3}) + \frac{1}{8} arctan (tan (\frac{θ}{2})) - (\frac{3}{8} arctan (\frac{tan ( \frac{θ}{2} )}{3}) - \frac{1}{8} arctan (tan (\frac{θ}{2})))) : 2 (\frac{1}{4} arctan (tan (\frac{θ}{2})) - \frac{5}{12} arctan (\frac{1}{3} tan (\frac{θ}{2})))

= 2 (\frac{1}{4} arctan (tan (\frac{θ}{2})) - \frac{5}{12} arctan (\frac{1}{3} tan (\frac{θ}{2})))

Add a constant to the solution

= 2 (\frac{1}{4} arctan (tan (\frac{θ}{2})) - \frac{5}{12} arctan (\frac{1}{3} tan (\frac{θ}{2}))) + C

\frac{d}{d y} (3 x^{2} y + e^{y})

\frac{3 x}{4}

\frac{\partial}{\partial x} ((4 x^{4} y^{2} + 9) 2)

\int - 4 sin^{2} (x + 2) d x

\int_{0}^{1} 2^{x} d x

What is the integral of (cos(θ))/(5+4cos(θ)) ?
The integral of (cos(θ))/(5+4cos(θ)) is 2(1/4 arctan(tan(θ/2))-5/12 arctan(1/3 tan(θ/2)))+C