What is the integral of 5xtan^2(x) ?

The integral of 5xtan^2(x) is 5(-x^2+xtan(x)+(x^2)/2+ln|cos(x)|)+C

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integral of 5xtan^2(x)

\int 5 x tan^{2} (x) d x

5 (- x^{2} + x tan (x) + \frac{x ^{2}}{2} + ln ∣ cos (x) ∣) + C

Solution steps

\int 5 x tan^{2} (x) d x

Take the constant out:

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

= 5 \cdot \int x tan^{2} (x) d x

Apply Integration By Parts

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

= 5 (x (- x + tan (x)) - \int - x + tan (x) d x)

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

= 5 (x (- x + tan (x)) - (- \frac{x ^{2}}{2} - ln ∣ cos (x) ∣))

Expand

x (- x + tan (x)) - (- \frac{x ^{2}}{2} - ln ∣ cos (x) ∣) : - x^{2} + x tan (x) + \frac{x ^{2}}{2} + ln ∣ cos (x) ∣

= 5 (- x^{2} + x tan (x) + \frac{x ^{2}}{2} + ln ∣ cos (x) ∣)

Add a constant to the solution

= 5 (- x^{2} + x tan (x) + \frac{x ^{2}}{2} + ln ∣ cos (x) ∣) + C

\int 2 x e^{- 4 x} d x

\int_{0}^{\infty} 7 e^{- x} d x

x \to 0 lim (2 x + 1)

\frac{d}{d x} (\frac{1}{2 x + 12})

\frac{\partial}{\partial x} (\frac{x \cdot d}{n \cdot L ( x , d )})

What is the integral of 5xtan^2(x) ?
The integral of 5xtan^2(x) is 5(-x^2+xtan(x)+(x^2)/2+ln|cos(x)|)+C