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

The integral of (2-cos(x)+sin(x))/(sin^2(x)) is 1/2 (3tan(x/2)+2ln|tan(x/2)|-cot(x/2))+C

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integral of (2-cos(x)+sin(x))/(sin^2(x))

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

\frac{3 u ^{2} + 2 u + 1}{u ^{2}} : 3 + \frac{2}{u} + \frac{1}{u ^{2}}

= \frac{1}{2} \cdot \int 3 + \frac{2}{u} + \frac{1}{u ^{2}} 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

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

\int 3 d u = 3 u

\int \frac{2}{u} d u = 2 ln ∣ u ∣

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

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

Substitute back

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

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

Use the basic trigonometric identity:

\frac{1}{tan ( x )} = cot (x)

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

Add a constant to the solution

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

\int \frac{15 x ^{2}}{( 9 - 2 x ^{3} ) ^{6}} d x

\int - e

\int - a

\int 3 x^{2} y e^{x^{3}} d x

\int (4 x^{2} + 5 x^{- 2}) d x

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