What is the integral of tan^2(xse)c^2x ?

The integral of tan^2(xse)c^2x is (c^2)/(2e^2s^2)(-e^2s^2x^2+2esxtan(esx)+2ln|cos(esx)|)+C

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integral of tan^2(xse)c^2x

\int tan^{2} (x se) c^{2} x d x

\frac{c ^{2}}{2 e ^{2} s ^{2}} (- e^{2} s^{2} x^{2} + 2 es x tan (es x) + 2 ln ∣ cos (es x) ∣) + C

Solution steps

\int tan^{2} (x se) 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 tan^{2} (se x) x d x

Apply u-substitution

: \int \frac{u tan ^{2} ( u )}{e ^{2} s ^{2}} d u

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = se x

= c^{2} \frac{1}{e ^{2} s ^{2}} (se x (- se x + tan (se x)) - (- \frac{( se x ) ^{2}}{2} - ln ∣ cos (se x) ∣))

Simplify

c^{2} \frac{1}{e ^{2} s ^{2}} (se x (- se x + tan (se x)) - (- \frac{( se x ) ^{2}}{2} - ln ∣ cos (se x) ∣)) : \frac{c ^{2}}{2 e ^{2} s ^{2}} (- e^{2} s^{2} x^{2} + 2 es x tan (es x) + 2 ln ∣ cos (es x) ∣)

= \frac{c ^{2}}{2 e ^{2} s ^{2}} (- e^{2} s^{2} x^{2} + 2 es x tan (es x) + 2 ln ∣ cos (es x) ∣)

Add a constant to the solution

= \frac{c ^{2}}{2 e ^{2} s ^{2}} (- e^{2} s^{2} x^{2} + 2 es x tan (es x) + 2 ln ∣ cos (es x) ∣) + C

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2 cos (2 t)

y^{^{''}} + 8 y^{^{'}} + 25 y = 0, y (3319) = - 3, y^{^{'}} (3319) = 1

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What is the integral of tan^2(xse)c^2x ?
The integral of tan^2(xse)c^2x is (c^2)/(2e^2s^2)(-e^2s^2x^2+2esxtan(esx)+2ln|cos(esx)|)+C