What is the integral of tan^2(xse)c^{10}x ?

The integral of tan^2(xse)c^{10}x is (c^{10})/(2e^2s^2)(-e^2s^2x^2+2esxtan(esx)+2ln|cos(esx)|)+C

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integral of tan^2(xse)c^{10}x

\int tan^{2} (x se) c^{10} x d x

\frac{c ^{10}}{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^{10} x d x

Take the constant out:

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

= c^{10} \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^{10} \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^{10} \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^{10} \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^{10} \frac{1}{e ^{2} s ^{2}} (u (- u + tan (u)) - (- \frac{u ^{2}}{2} - ln ∣ cos (u) ∣))

Substitute back

u = se x

= c^{10} \frac{1}{e ^{2} s ^{2}} (se x (- se x + tan (se x)) - (- \frac{( se x ) ^{2}}{2} - ln ∣ cos (se x) ∣))

Simplify

c^{10} \frac{1}{e ^{2} s ^{2}} (se x (- se x + tan (se x)) - (- \frac{( se x ) ^{2}}{2} - ln ∣ cos (se x) ∣)) : \frac{c ^{10}}{2 e ^{2} s ^{2}} (- e^{2} s^{2} x^{2} + 2 es x tan (es x) + 2 ln ∣ cos (es x) ∣)

= \frac{c ^{10}}{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 ^{10}}{2 e ^{2} s ^{2}} (- e^{2} s^{2} x^{2} + 2 es x tan (es x) + 2 ln ∣ cos (es x) ∣) + C

\int 2 1 - e^{- x} e^{- x} d x

- 5 cos (0.4 t)

y^{^{'}} = y^{2}

\int u \cdot e^{u} d u

x \frac{d y}{d x} = y + x^{2} sin (x)

What is the integral of tan^2(xse)c^{10}x ?
The integral of tan^2(xse)c^{10}x is (c^{10})/(2e^2s^2)(-e^2s^2x^2+2esxtan(esx)+2ln|cos(esx)|)+C