What is the integral of 7tan^2(xse)c^3x ?

The integral of 7tan^2(xse)c^3x is (7c^3)/(2e^2s^2)(-e^2s^2x^2+2esxtan(esx)+2ln|cos(esx)|)+C

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integral of 7tan^2(xse)c^3x

\int 7 tan^{2} (x se) c^{3} x

\frac{7 c ^{3}}{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 7 tan^{2} (x se) c^{3} x d x

Take the constant out:

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

= 7 c^{3} \cdot \int tan^{2} (se x) x d x

Apply u-substitution

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

= 7 c^{3} \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

= 7 c^{3} \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

= 7 c^{3} \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) ∣

= 7 c^{3} \frac{1}{e ^{2} s ^{2}} (u (- u + tan (u)) - (- \frac{u ^{2}}{2} - ln ∣ cos (u) ∣))

Substitute back

u = se x

= 7 c^{3} \frac{1}{e ^{2} s ^{2}} (se x (- se x + tan (se x)) - (- \frac{( se x ) ^{2}}{2} - ln ∣ cos (se x) ∣))

Simplify

7 c^{3} \frac{1}{e ^{2} s ^{2}} (se x (- se x + tan (se x)) - (- \frac{( se x ) ^{2}}{2} - ln ∣ cos (se x) ∣)) : \frac{7 c ^{3}}{2 e ^{2} s ^{2}} (- e^{2} s^{2} x^{2} + 2 es x tan (es x) + 2 ln ∣ cos (es x) ∣)

= \frac{7 c ^{3}}{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{7 c ^{3}}{2 e ^{2} s ^{2}} (- e^{2} s^{2} x^{2} + 2 es x tan (es x) + 2 ln ∣ cos (es x) ∣) + C

\frac{\partial}{\partial x} (\frac{x y}{x + y} + z)

\frac{d}{d x} (\frac{( 2 x + 3 ) ^{4}}{( 3 x - 2 ) ^{5}})

\int_{a}^{b} arctan (x) d x

4 y^{^{'}} - (\frac{2}{x}) y = (- \frac{3}{2}) x y^{\frac{5}{2}}

x \to 0 lim (\frac{sin ( x )}{7 x})

What is the integral of 7tan^2(xse)c^3x ?
The integral of 7tan^2(xse)c^3x is (7c^3)/(2e^2s^2)(-e^2s^2x^2+2esxtan(esx)+2ln|cos(esx)|)+C