What is the integral of (8x+3sin(x))^2 ?

The integral of (8x+3sin(x))^2 is (64x^3)/3+48(-xcos(x)+sin(x))+9/2 (x-1/2 sin(2x))+C

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integral of (8x+3sin(x))^2

\int (8 x + 3 sin (x))^{2} d x

\frac{64 x ^{3}}{3} + 48 (- x cos (x) + sin (x)) + \frac{9}{2} (x - \frac{1}{2} sin (2 x)) + C

Solution steps

\int (8 x + 3 sin (x))^{2} d x

Expand

(8 x + 3 sin (x))^{2} : 64 x^{2} + 48 x sin (x) + 9 sin^{2} (x)

= \int 64 x^{2} + 48 x sin (x) + 9 sin^{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 64 x^{2} d x + \int 48 x sin (x) d x + \int 9 sin^{2} (x) d x

\int 64 x^{2} d x = \frac{64 x ^{3}}{3}

\int 48 x sin (x) d x = 48 (- x cos (x) + sin (x))

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

= \frac{64 x ^{3}}{3} + 48 (- x cos (x) + sin (x)) + \frac{9}{2} (x - \frac{1}{2} sin (2 x))

Add a constant to the solution

= \frac{64 x ^{3}}{3} + 48 (- x cos (x) + sin (x)) + \frac{9}{2} (x - \frac{1}{2} sin (2 x)) + C

\int (\frac{1}{4 - 25 x ^{2}}) d x

\frac{14 x + 4 x ^{2}}{( 7 + 4 x ) ^{2}}

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

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

\frac{d y}{d x} = \frac{( 2 x ( y - 1 ) )}{( x ^{2} + 3 )}

What is the integral of (8x+3sin(x))^2 ?
The integral of (8x+3sin(x))^2 is (64x^3)/3+48(-xcos(x)+sin(x))+9/2 (x-1/2 sin(2x))+C