What is the integral of tan^2(xse)c^4x ?

The integral of tan^2(xse)c^4x is (c^4)/(2e^2s^2)(-e^2s^2x^2+2esxtan(esx)+2ln|cos(esx)|)+C

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integral of tan^2(xse)c^4x

\int tan^{2} (x se) c^{4} x d x

\frac{c ^{4}}{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^{4} x d x

Take the constant out:

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

= c^{4} \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^{4} \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^{4} \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^{4} \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^{4} \frac{1}{e ^{2} s ^{2}} (u (- u + tan (u)) - (- \frac{u ^{2}}{2} - ln ∣ cos (u) ∣))

Substitute back

u = se x

= c^{4} \frac{1}{e ^{2} s ^{2}} (se x (- se x + tan (se x)) - (- \frac{( se x ) ^{2}}{2} - ln ∣ cos (se x) ∣))

Simplify

c^{4} \frac{1}{e ^{2} s ^{2}} (se x (- se x + tan (se x)) - (- \frac{( se x ) ^{2}}{2} - ln ∣ cos (se x) ∣)) : \frac{c ^{4}}{2 e ^{2} s ^{2}} (- e^{2} s^{2} x^{2} + 2 es x tan (es x) + 2 ln ∣ cos (es x) ∣)

= \frac{c ^{4}}{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 ^{4}}{2 e ^{2} s ^{2}} (- e^{2} s^{2} x^{2} + 2 es x tan (es x) + 2 ln ∣ cos (es x) ∣) + C

\frac{d}{d x} (ln (3 x^{2} - 4 x))

x \to 2 lim (\frac{x ^{2}}{x ^{2} + 4})

\int cos (x) + \frac{1}{8} x d x

f^{^{''}} (x) = - 2 + 36 x - 12 x^{2}

- 4 sin (2 x)

What is the integral of tan^2(xse)c^4x ?
The integral of tan^2(xse)c^4x is (c^4)/(2e^2s^2)(-e^2s^2x^2+2esxtan(esx)+2ln|cos(esx)|)+C