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

The integral of x(3sec^2(x)+2tan^2(x)) is -x^2+5xtan(x)+5ln|cos(x)|+C

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

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

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

Solution steps

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

Rewrite using trig identities

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

Expand

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

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

= 6 (\frac{1}{2} x tan (x) + \frac{1}{2} ln ∣ cos (x) ∣) - x^{2} + 2 x tan (x) + 2 ln ∣ cos (x) ∣

Simplify

6 (\frac{1}{2} x tan (x) + \frac{1}{2} ln ∣ cos (x) ∣) - x^{2} + 2 x tan (x) + 2 ln ∣ cos (x) ∣ : - x^{2} + 5 x tan (x) + 5 ln ∣ cos (x) ∣

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

Add a constant to the solution

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

y = 3 x^{4}

\frac{d}{d t} (4 t)

y = 7 4 x^{3}

\frac{d}{d x} (2 x + cot (x))

3 y^{^{''}} + 2 y = x^{2} e^{- 3 x}

What is the integral of x(3sec^2(x)+2tan^2(x)) ?
The integral of x(3sec^2(x)+2tan^2(x)) is -x^2+5xtan(x)+5ln|cos(x)|+C