What is the integral of tan^2(2xse)c^22x ?

The integral of tan^2(2xse)c^22x is (c^2)/(2e^2s^2)(-2e^2s^2x^2+2esxtan(2esx)+ln|cos(2esx)|)+C

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integral of tan^2(2xse)c^22x

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{u tan ^{2} ( u )}{4 e ^{2} s ^{2}} d u

= 2 c^{2} \cdot \int \frac{u tan ^{2} ( u )}{4 e ^{2} s ^{2}} d u

Take the constant out:

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

= 2 c^{2} \frac{1}{4 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

= 2 c^{2} \frac{1}{4 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) ∣

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

Substitute back

u = 2 se x

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

Simplify

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

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

Add a constant to the solution

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

y^{^{'}} + y - 2 cos (x) = 0

t^{10} e^{- 7 t}

\int sin (\frac{nπ x}{3}) d x

n = 0 \sum \infty \frac{1}{n + 3}

\int \frac{1}{2 x ^{5}} d x

What is the integral of tan^2(2xse)c^22x ?
The integral of tan^2(2xse)c^22x is (c^2)/(2e^2s^2)(-2e^2s^2x^2+2esxtan(2esx)+ln|cos(2esx)|)+C