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+4xtan(x)+4ln|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} + 4 x tan (x) + 4 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} + 4 x tan (x) + 4 ln ∣ cos (x) ∣

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

Add a constant to the solution

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

\frac{d}{d x} (cos (5 x))

\int_{0}^{\frac{π}{2}} 5 cos^{5} (x) d x

\int_{0}^{1} 1 + (6 x)^{2} d x

\int \frac{x ^{2} - 4 x + 3}{x ^{3} + 2 x ^{2} + x} d x

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

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+4xtan(x)+4ln|cos(x)|+C