What is the integral of 1/((2+cos(x))^2) ?

The integral of 1/((2+cos(x))^2) is

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integral of 1/((2+cos(x))^2)

\int \frac{1}{( 2 + cos ( x ) ) ^{2}} d x

\frac{4}{3 3} arctan (\frac{1}{3} tan (\frac{x}{2})) + \frac{1}{6 3} sin (2 arctan (\frac{1}{3} tan (\frac{x}{2}))) - \frac{tan ( \frac{x}{2} )}{tan ^{2} ( \frac{x}{2} ) + 3} + C

Solution steps

\int \frac{1}{( 2 + cos ( x ) ) ^{2}} d x

Apply u-substitution

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

Expand

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

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

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

= 2 (\frac{1}{2 3} arctan (\frac{u}{3}) - \frac{u}{2 ( 3 + u ^{2} )} + \frac{1}{6 3} (arctan (\frac{u}{3}) + \frac{1}{2} sin (2 arctan (\frac{u}{3}))))

Substitute back

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

= 2 (\frac{1}{2 3} arctan (\frac{tan ( \frac{x}{2} )}{3}) - \frac{tan ( \frac{x}{2} )}{2 ( 3 + tan ^{2} ( \frac{x}{2} ) )} + \frac{1}{6 3} (arctan (\frac{tan ( \frac{x}{2} )}{3}) + \frac{1}{2} sin (2 arctan (\frac{tan ( \frac{x}{2} )}{3}))))

Simplify

2 (\frac{1}{2 3} arctan (\frac{tan ( \frac{x}{2} )}{3}) - \frac{tan ( \frac{x}{2} )}{2 ( 3 + tan ^{2} ( \frac{x}{2} ) )} + \frac{1}{6 3} (arctan (\frac{tan ( \frac{x}{2} )}{3}) + \frac{1}{2} sin (2 arctan (\frac{tan ( \frac{x}{2} )}{3})))) : \frac{4}{3 3} arctan (\frac{1}{3} tan (\frac{x}{2})) + \frac{1}{6 3} sin (2 arctan (\frac{1}{3} tan (\frac{x}{2}))) - \frac{tan ( \frac{x}{2} )}{tan ^{2} ( \frac{x}{2} ) + 3}

= \frac{4}{3 3} arctan (\frac{1}{3} tan (\frac{x}{2})) + \frac{1}{6 3} sin (2 arctan (\frac{1}{3} tan (\frac{x}{2}))) - \frac{tan ( \frac{x}{2} )}{tan ^{2} ( \frac{x}{2} ) + 3}

Add a constant to the solution

= \frac{4}{3 3} arctan (\frac{1}{3} tan (\frac{x}{2})) + \frac{1}{6 3} sin (2 arctan (\frac{1}{3} tan (\frac{x}{2}))) - \frac{tan ( \frac{x}{2} )}{tan ^{2} ( \frac{x}{2} ) + 3} + C

\int_{- 1}^{1} 1.5 x^{4} d x

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

t a y l or ln (5 - 2 x)

x \to \infty + lim (x^{12})

\frac{d}{d t} (sin (\frac{t}{2}))