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

The integral of (e^x-cos(x))^2 is 1/2 e^{2x}-e^xsin(x)-e^xcos(x)+1/2 x+1/4 sin(2x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (e^x-cos(x))^2

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

\frac{1}{2} e^{2 x} - e^{x} sin (x) - e^{x} cos (x) + \frac{1}{2} x + \frac{1}{4} sin (2 x) + C

Solution steps

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

Apply u-substitution

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

Expand

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

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

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

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

\int \frac{cos ^{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}{4} sin (2 x)

= \frac{1}{2} e^{2 x} - e^{x} sin (x) - e^{x} cos (x) + \frac{1}{2} x + \frac{1}{4} 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}{4} sin (2 x) + C

What is the integral of (e^x-cos(x))^2 ?
The integral of (e^x-cos(x))^2 is 1/2 e^{2x}-e^xsin(x)-e^xcos(x)+1/2 x+1/4 sin(2x)+C