# Thread: is this integrer solved properly?

1. ## is this integrer solved properly?

Let's consider the integrer:

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

Let's assume that:

$cos(x)+sin(x) = u$
$\Big (cos(x)-sin(x) \Big ) dx = du$
$\Bigg ( \Big (cos(x)-sin(x) \Big ) \frac{cos(x)+sin(x)}{cos(x)+sin(x)} \Bigg )dx = du$
$\frac{cos^{2}(x)-sin^{2}(x)}{cos(x)+sin(x)} dx = du$
$\frac{cos(2x)}{u} dx = du$
$cos(2x) dx = u du$

then our interger:

$- \int ue^udu$

next:

$f(x) = e^u, f'(x) = e^u$
oraz
$g(x) = u, g'(x) = 1$

and finally we get:

$- \int ue^udu = ue^u - \int e^u = ue^u - e^u +C = e^u(u - 1) + C$

so:

$- \int cos(2x)e^{cos(x)+sin(x)} = e^{cos(x)+sin(x)} \Big (cos(x)+sin(x)+1 \Big ) + C$

...and I am not sure if that solution is correct.

2. $-\int cos(2x)e^{cos(x)+sin(x)}=e^{cos(x)+sin(x)} \Big (cos(x)+sin(x)+1 \Big) +c " alt="-\int cos(2x)e^{cos(x)+sin(x)}=e^{cos(x)+sin(x)} \Big (cos(x)+sin(x)+1 \Big) +c " />

you should just change the red signs negative to positive and the positive to the negative

and this come from

sinx derive is cos and cosx derive -sinx

look carefully to this

3. $

((-))\int {cos(2x)e^{cos(x)+sin(x)}}=e^{cos(x)+sin(x)} \Big (cos(x)+sin(x)((+))1 \Big) +c

$

you should just change the sign in the arcs negative to positive and the positive to the negative