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

The integral of xtan^2(2x) is -x^2+1/2 xtan(2x)+(x^2)/2+1/4 ln|cos(2x)|+C

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integral of xtan^2(2x)

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

- x^{2} + \frac{1}{2} x tan (2 x) + \frac{x ^{2}}{2} + \frac{1}{4} ln ∣ cos (2 x) ∣ + C

Solution steps

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

Apply Integration By Parts

: x (- x + \frac{1}{2} tan (2 x)) - \int - x + \frac{1}{2} tan (2 x) d x

= x (- x + \frac{1}{2} tan (2 x)) - \int - x + \frac{1}{2} tan (2 x) d x

\int - x + \frac{1}{2} tan (2 x) d x = - \frac{x ^{2}}{2} - \frac{1}{4} ln ∣ cos (2 x) ∣

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

Simplify

x (- x + \frac{1}{2} tan (2 x)) - (- \frac{x ^{2}}{2} - \frac{1}{4} ln ∣ cos (2 x) ∣) : - x^{2} + \frac{1}{2} x tan (2 x) + \frac{x ^{2}}{2} + \frac{1}{4} ln ∣ cos (2 x) ∣

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

Add a constant to the solution

= - x^{2} + \frac{1}{2} x tan (2 x) + \frac{x ^{2}}{2} + \frac{1}{4} ln ∣ cos (2 x) ∣ + C

x^{4} e^{x}

\int (e^{y} + 1)^{- 2} e^{y} d y

\frac{\partial}{\partial x} (y ln (x^{4} + y^{3} + 2) - 3)

\frac{d ^{2}}{d x ^{2}} (sin (b x) - 2 cos (b x))

\int \frac{1}{b x} d x

What is the integral of xtan^2(2x) ?
The integral of xtan^2(2x) is -x^2+1/2 xtan(2x)+(x^2)/2+1/4 ln|cos(2x)|+C