Question

Evaluate the integral\(\int {{{\rm{e}}^{\rm{x}}}{\rm{cosxdx}}} \).

Short answer

The integral value of the given equation is\(\int {{{\rm{e}}^{\rm{x}}}} {\rm{cosxdx = }}\frac{{\rm{1}}}{{\rm{2}}}{{\rm{e}}^{\rm{x}}}{\rm{(cosx + sinx) + C}}\).

Step-by-step solution

Expand the equation.

\({\mathop{\rm I}\nolimits} = \int {{e^x}} cosxdx\)

Let\(\;{\rm{u = cosx}}\;\;\;{\rm{,dv = }}{{\rm{e}}^{\rm{x}}}{\rm{dx}}\)

Then \(\;\;\;{\rm{du = - sinxdx}}\;\;\;{\rm{,v = }}{{\rm{e}}^{\rm{x}}}\)

Known\(\int {\rm{u}} {\rm{dv = u}}{\rm{.v - }}\int {\rm{v}} {\rm{du}}\)

Parts integration yields

\(\int {{{\rm{e}}^{\rm{x}}}} {\rm{cosxdx = }}{{\rm{e}}^{\rm{x}}}{\rm{cosx + }}\int {{{\rm{e}}^{\rm{x}}}} {\rm{sinxdx}}\)

Evaluate the equation.

Parts integration once more

Let\(\;{\rm{u = sinx}}\;\;\;{\rm{,dv = }}{{\rm{e}}^{\rm{x}}}{\rm{dx}}\)

Then\({\rm{du = cosxdx}}\;\;\;{\rm{,v = }}{{\rm{e}}^{\rm{x}}}\)

Known\(\int {\rm{u}} {\rm{dv = u}}{\rm{.v - }}\int {\rm{v}} {\rm{du}}\)

Parts integration yields

\(\begin{aligned}{c}\int {{{\rm{e}}^{\rm{x}}}} {\rm{cosxdx = }}{{\rm{e}}^{\rm{x}}}{\rm{cosx + }}{{\rm{e}}^{\rm{x}}}{\rm{sinx - }}\int {{{\rm{e}}^{\rm{x}}}} {\rm{cosxdx}}\int {{{\rm{e}}^{\rm{x}}}} {\rm{cosxdx + }}\int {{{\rm{e}}^{\rm{x}}}} {\rm{cosxdx}}\\{\rm{ = }}{{\rm{e}}^{\rm{x}}}{\rm{cosx + }}{{\rm{e}}^{\rm{x}}}{\rm{sinx2}}\int {{{\rm{e}}^{\rm{x}}}} {\rm{cosxdx}}\\{\rm{ = }}{{\rm{e}}^{\rm{x}}}{\rm{cosx + }}{{\rm{e}}^{\rm{x}}}{\rm{sinx + c}}\int {{{\rm{e}}^{\rm{x}}}} {\rm{cosxdx}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left( {{{\rm{e}}^{\rm{x}}}{\rm{cosx + }}{{\rm{e}}^{\rm{x}}}{\rm{sinx}}} \right){\rm{ + C}}\int {{{\rm{e}}^{\rm{x}}}} {\rm{cosxdx}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}{{\rm{e}}^{\rm{x}}}{\rm{(cosx + sinx) + C}}\end{aligned}\)

Therefore, the integral value of the given equation is\(\int {{{\rm{e}}^{\rm{x}}}} {\rm{cosxdx = }}\frac{{\rm{1}}}{{\rm{2}}}{{\rm{e}}^{\rm{x}}}{\rm{(cosx + sinx) + C}}\).