What is the integral of x(3sec^2(x)-tan^2(x)) ?

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

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integral of x(3sec^2(x)-tan^2(x))

\int x (3 sec^{2} (x) - tan^{2} (x)) d x

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

Solution steps

\int x (3 sec^{2} (x) - tan^{2} (x)) d x

Rewrite using trig identities

= \int \frac{x ( 3 - sin ^{2} ( x ) )}{cos ^{2} ( x )} d x

Expand

\frac{x ( 3 - sin ^{2} ( x ) )}{cos ^{2} ( x )} : \frac{3 x}{cos ^{2} ( x )} - \frac{x sin ^{2} ( x )}{cos ^{2} ( x )}

= \int \frac{3 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{3 x}{cos ^{2} ( x )} d x - \int \frac{x sin ^{2} ( x )}{cos ^{2} ( x )} d x

\int \frac{3 x}{cos ^{2} ( x )} d x = 6 (\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) ∣

= 6 (\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

6 (\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} + 2 x tan (x) + 2 ln ∣ cos (x) ∣

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

Add a constant to the solution

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

\int \frac{x}{x ^{2} + 6 x + 18} d x

\frac{s + 1}{s ( s ^{2} + 4 s + 4 )}

\frac{d}{d x} (x csc (x))

x^{3} + 2 x

\int 9 sin (x) cos (x) d x

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