What is the integral of cot^2(xcs)c^2x ?

The integral of cot^2(xcs)c^2x is -c^2x^2-c/s xcot(csx)+(c^2x^2)/2+1/(s^2)ln|sin(csx)|+C

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integral of cot^2(xcs)c^2x

\int cot^{2} (x cs) c^{2} x

- c^{2} x^{2} - \frac{c}{s} x cot (cs x) + \frac{c ^{2} x ^{2}}{2} + \frac{1}{s ^{2}} ln ∣ sin (cs x) ∣ + C

Solution steps

\int cot^{2} (x cs) c^{2} x d x

Take the constant out:

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

= c^{2} \cdot \int cot^{2} (cs x) x d x

Apply u-substitution

: \int \frac{u cot ^{2} ( u )}{c ^{2} s ^{2}} d u

= c^{2} \cdot \int \frac{u cot ^{2} ( u )}{c ^{2} s ^{2}} d u

Take the constant out:

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

= c^{2} \frac{1}{c ^{2} s ^{2}} \cdot \int u cot^{2} (u) d u

Apply Integration By Parts

: u (- u - cot (u)) - \int - u - cot (u) d u

= c^{2} \frac{1}{c ^{2} s ^{2}} (u (- u - cot (u)) - \int - u - cot (u) d u)

\int - u - cot (u) d u = - \frac{u ^{2}}{2} - ln ∣ sin (u) ∣

= c^{2} \frac{1}{c ^{2} s ^{2}} (u (- u - cot (u)) - (- \frac{u ^{2}}{2} - ln ∣ sin (u) ∣))

Substitute back

u = cs x

= c^{2} \frac{1}{c ^{2} s ^{2}} (cs x (- cs x - cot (cs x)) - (- \frac{( cs x ) ^{2}}{2} - ln ∣ sin (cs x) ∣))

Simplify

c^{2} \frac{1}{c ^{2} s ^{2}} (cs x (- cs x - cot (cs x)) - (- \frac{( cs x ) ^{2}}{2} - ln ∣ sin (cs x) ∣)) : - c^{2} x^{2} - \frac{c}{s} x cot (cs x) + \frac{c ^{2} x ^{2}}{2} + \frac{1}{s ^{2}} ln ∣ sin (cs x) ∣

= - c^{2} x^{2} - \frac{c}{s} x cot (cs x) + \frac{c ^{2} x ^{2}}{2} + \frac{1}{s ^{2}} ln ∣ sin (cs x) ∣

Add a constant to the solution

= - c^{2} x^{2} - \frac{c}{s} x cot (cs x) + \frac{c ^{2} x ^{2}}{2} + \frac{1}{s ^{2}} ln ∣ sin (cs x) ∣ + C

f (x) = x - 3

\int - cos (t) sin (t) d t

(\frac{1}{X})^{^{'}}

\int_{- 2}^{2} x (x^{2} + 1)^{3} d x

\frac{d}{d x} (\frac{( 2 x - 3 ) ^{- 3}}{( 2 - 3 x ) ^{2}})

What is the integral of cot^2(xcs)c^2x ?
The integral of cot^2(xcs)c^2x is -c^2x^2-c/s xcot(csx)+(c^2x^2)/2+1/(s^2)ln|sin(csx)|+C