What is the integral of [5sin^2(3x)+tan(2x)] ?

The integral of [5sin^2(3x)+tan(2x)] is 5/2 (x-1/6 sin(6x))-1/2 ln|cos(2x)|+C

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integral of [5sin^2(3x)+tan(2x)]

\int [5 sin^{2} (3 x) + tan (2 x)] d x

\frac{5}{2} (x - \frac{1}{6} sin (6 x)) - \frac{1}{2} ln ∣ cos (2 x) ∣ + C

Solution steps

\int 5 sin^{2} (3 x) + tan (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 5 sin^{2} (3 x) d x + \int tan (2 x) d x

\int 5 sin^{2} (3 x) d x = \frac{5}{2} (x - \frac{1}{6} sin (6 x))

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

= \frac{5}{2} (x - \frac{1}{6} sin (6 x)) - \frac{1}{2} ln ∣ cos (2 x) ∣

Add a constant to the solution

= \frac{5}{2} (x - \frac{1}{6} sin (6 x)) - \frac{1}{2} ln ∣ cos (2 x) ∣ + C

\frac{d}{d x} (4 x + 8 + \frac{14400}{x})

\int 2 x + 4 e^{x} d x

\int_{0}^{1} \frac{x}{2} d x

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

\frac{d}{d x} (3 sin^{5} (x))

What is the integral of [5sin^2(3x)+tan(2x)] ?
The integral of [5sin^2(3x)+tan(2x)] is 5/2 (x-1/6 sin(6x))-1/2 ln|cos(2x)|+C