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

The integral of (e^x+sin(x))^2 is 1/2 e^{2x}+e^xsin(x)-e^xcos(x)+1/2 (x-1/2 sin(2x))+C

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integral of (e^x+sin(x))^2

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

\frac{1}{2} e^{2 x} + e^{x} sin (x) - e^{x} cos (x) + \frac{1}{2} (x - \frac{1}{2} sin (2 x)) + C

Solution steps

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

Apply u-substitution

: \int \frac{( u + sin ( ln ( u ) ) ) ^{2}}{u} d u

= \int \frac{( u + sin ( ln ( u ) ) ) ^{2}}{u} d u

Expand

\frac{( u + sin ( ln ( u ) ) ) ^{2}}{u} : u + 2 sin (ln (u)) + \frac{sin ^{2} ( ln ( u ) )}{u}

= \int u + 2 sin (ln (u)) + \frac{sin ^{2} ( ln ( u ) )}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int u d u + \int 2 sin (ln (u)) d u + \int \frac{sin ^{2} ( ln ( u ) )}{u} d u

\int u d u = \frac{u ^{2}}{2}

\int 2 sin (ln (u)) d u = u sin (ln (u)) - u cos (ln (u))

\int \frac{sin ^{2} ( ln ( u ) )}{u} d u = \frac{1}{2} (ln (u) - \frac{1}{2} sin (2 ln (u)))

= \frac{u ^{2}}{2} + u sin (ln (u)) - u cos (ln (u)) + \frac{1}{2} (ln (u) - \frac{1}{2} sin (2 ln (u)))

Substitute back

u = e^{x}

= \frac{( e ^{x} ) ^{2}}{2} + e^{x} sin (ln (e^{x})) - e^{x} cos (ln (e^{x})) + \frac{1}{2} (ln (e^{x}) - \frac{1}{2} sin (2 ln (e^{x})))

Simplify

\frac{( e ^{x} ) ^{2}}{2} + e^{x} sin (ln (e^{x})) - e^{x} cos (ln (e^{x})) + \frac{1}{2} (ln (e^{x}) - \frac{1}{2} sin (2 ln (e^{x}))) : \frac{1}{2} e^{2 x} + e^{x} sin (x) - e^{x} cos (x) + \frac{1}{2} (x - \frac{1}{2} sin (2 x))

= \frac{1}{2} e^{2 x} + e^{x} sin (x) - e^{x} cos (x) + \frac{1}{2} (x - \frac{1}{2} sin (2 x))

Add a constant to the solution

= \frac{1}{2} e^{2 x} + e^{x} sin (x) - e^{x} cos (x) + \frac{1}{2} (x - \frac{1}{2} sin (2 x)) + C

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What is the integral of (e^x+sin(x))^2 ?
The integral of (e^x+sin(x))^2 is 1/2 e^{2x}+e^xsin(x)-e^xcos(x)+1/2 (x-1/2 sin(2x))+C