Question

Integrate the following problem:

Answer 1

Answer:

$\displaystyle \frac{2 \cdot sin2x-cos2x}{5e^x} + C$

Step-by-step explanation:

The integration by parts formula is: $\displaystyle \int udv = uv - \int vdu$

Let's find u, du, dv, and v for $\displaystyle \int e^-^x \cdot cos2x \ dx$ .

• $u=e^-^x$
• $du=-e^-^x dx$

• $dv=cos2x \ dx$
• $v= \frac{sin2x}{2}$

Plug these values into the IBP formula:

• $\displaystyle \int e^-^x \cdot cos2x \ dx = e^-^x \cdot \frac{sin2x}{2} - \int \frac{sin2x}{2} \cdot -e^-^x dx$
• $\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \int \frac{sin2x}{2} \cdot -e^-^x dx$

Now let's evaluate the integral $\displaystyle \int \frac{sin2x}{2} \cdot -e^-^x dx$.

Let's find u, du, dv, and v for this integral:

• $u=-e^-^x$
• $du=e^-^x dx$
• $dv=\frac{sin2x}{2} dx$
• $v=\frac{-cos2x}{4}$  

Plug these values into the IBP formula:

• $\displaystyle \int -e^-^x \cdot \frac{sin2x}{x}dx = -e^-^x \cdot \frac{-cos2x}{4} - \int \frac{-cos2x}{4}\cdot e^-^x dx$

Factor 1/4 out of the integral and we are left with the exact same integral from the question.

• $\displaystyle \int -e^-^x \cdot \frac{sin2x}{x}dx = -e^-^x \cdot \frac{-cos2x}{4} + \frac{1}{4} \int cos2x \cdot e^-^x dx$

Let's substitute this back into the first IBP equation.

• $\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \Big [ -e^-^x \cdot \frac{-cos2x}{4} + \frac{1}{4} \int cos2x \cdot e^-^x dx \Big ]$  

Simplify inside the brackets.

• $\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \Big [ \frac{e^-^x \cdot cos2x}{4} + \frac{1}{4} \int cos2x \cdot e^-^x dx \Big ]$

Distribute the negative sign into the parentheses.

• $\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \frac{e^-^x \cdot cos2x}{4} - \frac{1}{4} \int cos2x \cdot e^-^x dx$

Add the like term to the left side.

• $\displaystyle \int e^-^x \cdot cos2x \ dx + \frac{1}{4} \int cos2x \cdot e^-^x dx= \frac{e^-^x sin2x}{2} - \frac{e^-^x \cdot cos2x}{4}$  
• $\displaystyle \frac{5}{4} \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \frac{e^-^x \cdot cos2x}{4}$  

Make the fractions have common denominators.

• $\displaystyle \frac{5}{4} \int e^-^x \cdot cos2x \ dx = \frac{2e^-^x sin2x}{4} - \frac{e^-^x \cdot cos2x}{4}$

Simplify this equation.

• $\displaystyle \frac{5}{4} \int e^-^x \cdot cos2x \ dx = \frac{2e^-^x sin2x - e^-^x cos2x}{4}$

Multiply the right side by the reciprocal of 5/4.

• $\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{2e^-^x sin2x - e^-^x cos2x}{4} \cdot \frac{4}{5}$

The 4's cancel out and we are left with:

• $\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{2e^-^x sin2x - e^-^x cos2x}{5}$

Factor $e^-^x$ out of the numerator.

• $\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x(2 \cdot sin2x-cos2x)}{5}$

Simplify this by using exponential properties.

• $\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{2 \cdot sin2x-cos2x}{5e^x}$

The final answer is $\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{2 \cdot sin2x-cos2x}{5e^x} + C$.

Answer 2

Answer:

$\displaystyle\int e^{-x}\cos(2x)\, dx=\frac{2\sin(2x)-\cos(2x)}{5e^x}+C$

Step-by-step explanation:

We would like to integrate the following integral:

$\displaystyle \int e^{-x}\cdot \cos(2x)\, dx$

Since this is a product of two functions, we can consider using Integration by Parts given by:

$\displaystyle \int u\, dv =uv-\int v\, du$

So, let’s choose our u and dv. We can choose u base on the following guidelines: LIATE; or, logarithmic, inverse trig., algebraic, trigonometric, and exponential.

Since trigonometric comes before exponential, we will let:

$u=\cos(2x)\text{ and } dv=e^{-x}\, dx$

By finding the differential of the left and integrating the right, we acquire:

$du=-2\sin(2x)\text{ and } v=-e^{-x}$

So, our integral becomes:

$\displaystyle \int e^{-x}\cdot \cos(2x)\, dx=(\cos(2x))(-e^{-x})-\int (-e^{-x})(-2\sin(2x))\, dx$

Simplify:

$\displaystyle \int e^{-x}\cdot \cos(2x)\, dx=-e^{-x}\cos(2x)-2\int e^{-x}\sin(2x)\, dx$

Since we ended up with another integral of a product of two functions, we can apply integration by parts again. Using the above guidelines, we get that:

$u=\sin(2x)\text{ and } dv=e^{-x}\, dx$

By finding the differential of the left and integrating the right, we acquire:

$du=2\cos(2x)\, dx\text{ and } v=-e^{-x}$

This yields:

$\displaystyle \int e^{-x}\cdot \cos(2x)\, dx=-e^{-x}\cos(2x)-2\Big[(\sin(2x))(-e^{-x})-\int (-e^{-x})(2\sin(2x))\, dx\Big]$

Simplify:

$\displaystyle \int e^{-x}\cdot \cos(2x)\, dx=-e^{-x}\cos(2x)-2\Big[-e^{-x}\sin(2x)+2\int e^{-x}\cos(2x)\, dx\Big]$

We can distribute:

$\displaystyle \int e^{-x}\cdot \cos(2x)\, dx=-e^{-x}\cos(2x)+2e^{-x}\sin(2x)-4\int e^{-x}\cos(2x)\, dx$

The integral on the right is the same as our original integral. So, we can isolate it:

$\displaystyle \Big(\int e^{-x}\cos(2x)\, dx\Big)+4\Big(\int e^{-x}\cos(2x)\, dx)\Big)=-e^{-x}\cos(2x)+2e^{-x}\sin(2x)$

Combine like integrals:

$\displaystyle 5 \int e^{-x}\cos(2x)\, dx=-e^{-x}\cos(2x)+2e^{-x}\sin(2x)$

We can factor out an e⁻ˣ from the right:

$\displaystyle 5\int e^{-x}\cos(2x)\, dx=e^{-x}\Big(-\cos(2x)+2\sin(2x)\Big)$

Dividing both sides by 5 yields:

$\displaystyle \int e^{-x}\cos(2x)\, dx=\frac{e^{-x}}{5}\Big(-\cos(2x)+2\sin(2x)\Big)$

Rewrite. We of course also need the constant of integration. Therefore, our final answer is:

$\displaystyle\int e^{-x}\cos(2x)\, dx=\frac{2\sin(2x)-\cos(2x)}{5e^x}+C$