What is the integral of cot^2(θ)csc(θ) ?

The integral of cot^2(θ)csc(θ) is 1/8 (-cot^2(θ/2)-4ln|tan(θ/2)|+tan^2(θ/2))+C

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integral of cot^2(θ)csc(θ)

\int cot^{2} (θ) csc (θ) d θ

\frac{1}{8} (- cot^{2} (\frac{θ}{2}) - 4 ln tan (\frac{θ}{2}) + tan^{2} (\frac{θ}{2})) + C

Solution steps

\int cot^{2} (θ) csc (θ) d θ

Apply u-substitution

: \int \frac{( 1 - u ^{2} ) ^{2}}{4 u ^{3}} d u

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

Take the constant out:

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

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

Expand

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

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

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

\int \frac{2}{u} d u = 2 ln ∣ u ∣

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

= \frac{1}{4} (- \frac{1}{2 u ^{2}} - 2 ln ∣ u ∣ + \frac{u ^{2}}{2})

Substitute back

u = tan (\frac{θ}{2})

= \frac{1}{4} (- \frac{1}{2 tan ^{2} ( \frac{θ}{2} )} - 2 ln tan (\frac{θ}{2}) + \frac{tan ^{2} ( \frac{θ}{2} )}{2})

Simplify

\frac{1}{4} (- \frac{1}{2 tan ^{2} ( \frac{θ}{2} )} - 2 ln tan (\frac{θ}{2}) + \frac{tan ^{2} ( \frac{θ}{2} )}{2}) : \frac{1}{8} (- cot^{2} (\frac{θ}{2}) - 4 ln tan (\frac{θ}{2}) + tan^{2} (\frac{θ}{2}))

= \frac{1}{8} (- cot^{2} (\frac{θ}{2}) - 4 ln tan (\frac{θ}{2}) + tan^{2} (\frac{θ}{2}))

Add a constant to the solution

= \frac{1}{8} (- cot^{2} (\frac{θ}{2}) - 4 ln tan (\frac{θ}{2}) + tan^{2} (\frac{θ}{2})) + C

\frac{s - 1}{( s + 1 ) ( s ^{2} + 1 )}

f (x) = \frac{x - 8}{3 x - 4}, at x = 2

\frac{d}{d x} (cos (x z (x)))

\int x^{\frac{4}{3}} - 3 x^{\frac{5}{2}} d x

\int \frac{x ^{3}}{x + 1} d x

What is the integral of cot^2(θ)csc(θ) ?
The integral of cot^2(θ)csc(θ) is 1/8 (-cot^2(θ/2)-4ln|tan(θ/2)|+tan^2(θ/2))+C