What is the integral of x(2sec^2(x)+tan^2(x)) ?

The integral of x(2sec^2(x)+tan^2(x)) is -x^2+(x^2)/2+3xtan(x)+3ln|cos(x)|+C

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integral of x(2sec^2(x)+tan^2(x))

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

- x^{2} + \frac{x ^{2}}{2} + 3 x tan (x) + 3 ln ∣ cos (x) ∣ + C

Solution steps

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

Rewrite using trig identities

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

Expand

\frac{x ( 2 + sin ^{2} ( x ) )}{cos ^{2} ( x )} : \frac{2 x}{cos ^{2} ( x )} + \frac{x sin ^{2} ( x )}{cos ^{2} ( x )}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Simplify

4 (\frac{1}{2} x tan (x) + \frac{1}{2} ln ∣ cos (x) ∣) - x^{2} + x tan (x) + \frac{x ^{2}}{2} + ln ∣ cos (x) ∣ : - x^{2} + \frac{x ^{2}}{2} + 3 x tan (x) + 3 ln ∣ cos (x) ∣

= - x^{2} + \frac{x ^{2}}{2} + 3 x tan (x) + 3 ln ∣ cos (x) ∣

Add a constant to the solution

= - x^{2} + \frac{x ^{2}}{2} + 3 x tan (x) + 3 ln ∣ cos (x) ∣ + C

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 4

x \to \frac{π}{4} lim (\frac{sin ( x )}{x})

\frac{d y}{d t} = 2 y

\frac{d}{d y} (4 - y^{2})

\int_{- 7}^{0} (4 + 49 - x^{2}) d x

What is the integral of x(2sec^2(x)+tan^2(x)) ?
The integral of x(2sec^2(x)+tan^2(x)) is -x^2+(x^2)/2+3xtan(x)+3ln|cos(x)|+C