Solve ∫ e^xcosx wrt x=1/2e^x(cosx+si

$$\int e ^ { x } \cos x d x = \frac { 1 } { 2 } e ^ { x } ( \cos x + s i$$

Solve for s

$$s=-\frac{2iС}{e^{x}}-i\sin(x)$$

Steps for Solving Linear Equation

Use the distributive property to multiply \frac{1}{2}e^{x} by \cos(x)+si.

$$\int e^{x}\cos(x)\mathrm{d}x=\frac{1}{2}e^{x}\cos(x)+\frac{1}{2}ie^{x}s$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{1}{2}e^{x}\cos(x)+\frac{1}{2}ie^{x}s=\int e^{x}\cos(x)\mathrm{d}x$$

Subtract \frac{1}{2}e^{x}\cos(x) from both sides.

$$\frac{1}{2}ie^{x}s=\int e^{x}\cos(x)\mathrm{d}x-\frac{1}{2}e^{x}\cos(x)$$

The equation is in standard form.

$$\frac{ie^{x}}{2}s=\int \cos(x)e^{x}\mathrm{d}x-\frac{\cos(x)e^{x}}{2}$$

Divide both sides by \frac{1}{2}ie^{x}.

$$\frac{2\times \frac{ie^{x}}{2}s}{ie^{x}}=\frac{2\left(-\frac{\cos(x)e^{x}}{2}+\left(\frac{1}{4}-\frac{1}{4}i\right)e^{i\ln(e^{x})+x}+\left(\frac{1}{4}+\frac{1}{4}i\right)e^{-i\ln(e^{x})+x}+С\right)}{ie^{x}}$$

Dividing by \frac{1}{2}ie^{x} undoes the multiplication by \frac{1}{2}ie^{x}.

$$s=\frac{2\left(-\frac{\cos(x)e^{x}}{2}+\left(\frac{1}{4}-\frac{1}{4}i\right)e^{i\ln(e^{x})+x}+\left(\frac{1}{4}+\frac{1}{4}i\right)e^{-i\ln(e^{x})+x}+С\right)}{ie^{x}}$$

Divide \left(\frac{1}{4}+\frac{1}{4}i\right)e^{x-i\ln(e^{x})}+\left(\frac{1}{4}-\frac{1}{4}i\right)e^{x+i\ln(e^{x})}+С-\frac{e^{x}\cos(x)}{2} by \frac{1}{2}ie^{x}.

$$s=\frac{\left(-1-i\right)e^{i\ln(e^{x})}+\left(1-i\right)e^{-i\ln(e^{x})}+\frac{2С}{e^{x}}+2i\cos(x)}{2}$$