What is the integral of (2sin(x)-3cos(x))^2 ?

The integral of (2sin(x)-3cos(x))^2 is 13/2 x+5/4 sin(2x)-6sin^2(x)+C

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integral of (2sin(x)-3cos(x))^2

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

\frac{13}{2} x + \frac{5}{4} sin (2 x) - 6 sin^{2} (x) + C

Solution steps

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

Expand

(2 sin (x) - 3 cos (x))^{2} : 4 sin^{2} (x) - 12 sin (x) cos (x) + 9 cos^{2} (x)

= \int 4 sin^{2} (x) - 12 sin (x) cos (x) + 9 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 4 sin^{2} (x) d x - \int 12 sin (x) cos (x) d x + \int 9 cos^{2} (x) d x

\int 4 sin^{2} (x) d x = 2 x - sin (2 x)

\int 12 sin (x) cos (x) d x = 6 sin^{2} (x)

\int 9 cos^{2} (x) d x = \frac{9}{2} (x + \frac{1}{2} sin (2 x))

= 2 x - sin (2 x) - 6 sin^{2} (x) + \frac{9}{2} (x + \frac{1}{2} sin (2 x))

Simplify

2 x - sin (2 x) - 6 sin^{2} (x) + \frac{9}{2} (x + \frac{1}{2} sin (2 x)) : \frac{13}{2} x + \frac{5}{4} sin (2 x) - 6 sin^{2} (x)

= \frac{13}{2} x + \frac{5}{4} sin (2 x) - 6 sin^{2} (x)

Add a constant to the solution

= \frac{13}{2} x + \frac{5}{4} sin (2 x) - 6 sin^{2} (x) + C

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

\frac{d y}{d x} = \frac{- y - 6 x ^{2}}{x + 3}

x \to 3 - lim (h (x))

f (x) = 6 x

\int_{0}^{1} \frac{x + 1}{x ^{2} + 2 x - 3} d x

What is the integral of (2sin(x)-3cos(x))^2 ?
The integral of (2sin(x)-3cos(x))^2 is 13/2 x+5/4 sin(2x)-6sin^2(x)+C