What is the integral of tan^2(xco)s^3x ?

The integral of tan^2(xco)s^3x is (s^3)/(2c^2o^2)(-c^2o^2x^2+2coxtan(cox)+2ln|cos(cox)|)+C

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integral of tan^2(xco)s^3x

\int tan^{2} (x co) s^{3} x d x

\frac{s ^{3}}{2 c ^{2} o ^{2}} (- c^{2} o^{2} x^{2} + 2 co x tan (co x) + 2 ln ∣ cos (co x) ∣) + C

Solution steps

\int tan^{2} (x co) s^{3} x d x

Take the constant out:

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

= s^{3} \cdot \int tan^{2} (co x) x d x

Apply u-substitution

: \int \frac{u tan ^{2} ( u )}{c ^{2} o ^{2}} d u

= s^{3} \cdot \int \frac{u tan ^{2} ( u )}{c ^{2} o ^{2}} d u

Take the constant out:

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

= s^{3} \frac{1}{c ^{2} o ^{2}} \cdot \int u tan^{2} (u) d u

Apply Integration By Parts

: u (- u + tan (u)) - \int - u + tan (u) d u

= s^{3} \frac{1}{c ^{2} o ^{2}} (u (- u + tan (u)) - \int - u + tan (u) d u)

\int - u + tan (u) d u = - \frac{u ^{2}}{2} - ln ∣ cos (u) ∣

= s^{3} \frac{1}{c ^{2} o ^{2}} (u (- u + tan (u)) - (- \frac{u ^{2}}{2} - ln ∣ cos (u) ∣))

Substitute back

u = co x

= s^{3} \frac{1}{c ^{2} o ^{2}} (co x (- co x + tan (co x)) - (- \frac{( co x ) ^{2}}{2} - ln ∣ cos (co x) ∣))

Simplify

s^{3} \frac{1}{c ^{2} o ^{2}} (co x (- co x + tan (co x)) - (- \frac{( co x ) ^{2}}{2} - ln ∣ cos (co x) ∣)) : \frac{s ^{3}}{2 c ^{2} o ^{2}} (- c^{2} o^{2} x^{2} + 2 co x tan (co x) + 2 ln ∣ cos (co x) ∣)

= \frac{s ^{3}}{2 c ^{2} o ^{2}} (- c^{2} o^{2} x^{2} + 2 co x tan (co x) + 2 ln ∣ cos (co x) ∣)

Add a constant to the solution

= \frac{s ^{3}}{2 c ^{2} o ^{2}} (- c^{2} o^{2} x^{2} + 2 co x tan (co x) + 2 ln ∣ cos (co x) ∣) + C

x \to - 2 lim (\frac{- 3}{x + 2})

\int e^{x + 3 e^{x}} d x

\frac{d ^{2} y}{d x ^{2}} - 9 y = 0

\frac{\partial}{\partial x} (a - b x - \frac{c}{y} + d \frac{x}{y})

x \to 2 lim (2 x^{2} - x + 2)

What is the integral of tan^2(xco)s^3x ?
The integral of tan^2(xco)s^3x is (s^3)/(2c^2o^2)(-c^2o^2x^2+2coxtan(cox)+2ln|cos(cox)|)+C