Question

How to integrate `int e^x sinx cosx dx` ?

Answer 1

`int\ e^xsinxcosx\ dx=e^x/10sin(2x)-e^x/5cos(2x)+C`

Explanation:

First we can use the identity:

`2sinthetacostheta=sin2x`

which gives:

`int\ e^xsinxcosx\ dx=1/2int\ e^xsin(2x)\ dx`

Now we can use integration by parts. The formula is:

`int\ f(x)g'(x)\ dx=f(x)g(x)-int\ f'(x)g(x)\ dx`

I will let `f(x)=sin(2x)` and `g'(x)=e^x/2`. Applying the formula, we get:

`int\ e^x/2sin(2x)\ dx=sin(2x)e^x/2-int\ cos(2x)e^x\ dx`

Now we can apply integration by parts once more, this time with `f(x)=cos(2x)` and `g'(x)=e^x`:

`int\ e^x/2sin(2x)\ dx=sin(2x)e^x/2-(cos(2x)e^x-int\ -2sin(2x)e^x\ dx)`

`1/2int\ e^xsin(2x)\ dx=sin(2x)e^x/2-cos(2x)e^x-2int\ sin(2x)e^x\ dx`

Now we have the integral on both sides of the equality, so we can solve it like an equation. First, we add 2 times the integral to both sides:

`5/2int\ e^xsin(2x)\ dx=sin(2x)e^x/2-cos(2x)e^x+C`

Since we wanted a half as the coefficient on the original integral, we divide both sides by `5`:

`1/2int\ e^xsin(2x)\ dx=1/5(sin(2x)e^x/2-cos(2x)e^x)+C=`

`=e^x/10sin(2x)-e^x/5cos(2x)+C`

Answer 2

`int \ e^x \ sinxcosx \ dx = 1/10{e^x \ sin2x -2\ e^x \ cos2x} + C`

Explanation:

We seek:

`I = int \ e^x \ sinxcosx \ dx`

Which using the identity:

`sin 2x -= 2sinxcosx`

We can write as:

`I = 1/2 \ int \ e^x \ sin2x \ dx`
`I = 1/2 \ I_S`

Where for convenience we denote:

`I_S = int \ e^x \ sin2x \ dx`, and `I_C = int \ e^x \ cos2x \ dx`

We can apply integration by parts. Typically when assigning values of `u` and `dv`, we want to choose a function for `u` that will get simpler as we differentiate it. However, we see that `e^(x)` remain unchanged (being an exponential function) and `sinx` will alternate with `cosx`x as we differentiate or integrate.

In fact, we see it doesn't really matter which we choose for `u` and which for `dv`. So arbitarlity:

Let # { (u,=e^x, => (du)/dx,=e^x), ((dv)/dx,=sin2x, => v,=-1/2 cos2x ) :}#

Then plugging into the IBP formula:

`int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx`

We get:

`int \ (e^x)(sin2x) \ dx = (e^x)(-1/2cos2x) - int \ (-1/2cos2x)(e^x) \ dx`

`:. I_S = -1/2 \ e^x \ cos2x + 1/2 int \ e^x \ cos2x \ dx`
`:. I_S = -1/2 \ e^x \ cos2x + 1/2 I_C` ..... [A]

Now, we perform integration by parts once more.

Let # { (u,=e^x, => (du)/dx,=e^x), ((dv)/dx,=cos2x, => v,=1/2 sin2x ) :}#

Then plugging into the IBP formula we get:

`int \ (e^x)(cos2x) \ dx = (e^x)(1/2cos2x) - int \ (1/2sin2x)(e^x) \ dx`

`:. I_C = 1/2 \ e^x \ sin2x - 1/2 \ int e^x \ sin2x \ dx`
`:. I_C = 1/2 \ e^x \ sin2x - 1/2 I_S` ..... [B}

Now, we have two simultaneous equation in two unknowns `I_S`. and `I_C`, so substituting [B[ into [A] we have:

`I_S = -1/2 \ e^x \ cos2x + 1/2 {1/2 \ e^x \ sin2x - 1/2 I_S}`

`\ \ \ = -1/2 \ e^x \ cos2x + 1/4 \ e^x \ sin2x - 1/4 I_S`

`:. 5/4I_S = 1/4 \ e^x \ sin2x -1/2 \ e^x \ cos2x`

`:. I_S = 4/5{1/4 \ e^x \ sin2x -1/2 \ e^x \ cos2x}`

Leading to:

`I = 1/2 \ I_S + C`

`\ \ = 2/5{1/4 \ e^x \ sin2x -1/2 \ e^x \ cos2x} + C`

`\ \ = 1/10{e^x \ sin2x -2\ e^x \ cos2x} + C`