Question

integration
e^-xcos(e^-x)​

Answer 1

Given Integrand,

$\displaystyle \sf \int {e}^{ - x} cos( {e}^{ - x} )dx$

Let u = 1/e^x.

Differentiating both sides w.r.t x,

$\implies \sf \dfrac{du}{dx} = - {e}^{ - x} \\ \\ \implies \sf \: dx = - \dfrac{du}{ {e}^{ - x} }$

Now,

$\longrightarrow \displaystyle \sf - \int {e}^{ - x} cos(u) \dfrac{du}{ {e}^{ - x} } \\ \\ \longrightarrow \displaystyle \sf - \int cos(u)du \\ \\ \longrightarrow \displaystyle \sf - sin(u) + c \\ \\ \longrightarrow \displaystyle \sf - sin( {e}^{ - x} ) + c$

Thus,

$\boxed{ \boxed{ \displaystyle \sf \int {e}^{ - x} cos( {e}^{ - x} )dx = - sin( {e}^{ - x} ) + c}}$

Answer 2

Solution

Given :-

$\tt \implies \int {e}^{ - x} cos( {e}^{ - x})dx$

Using Substitution Method

so, Let

$\tt \implies \: {e}^{ - x} = t$

Now Differentiate Both the side with Respect to x , We get

$\tt \implies \: \dfrac{d( {e}^{ - x}) }{dx} = \dfrac{dt}{dx}$

$\tt \implies \: - e ^{ - x} = \dfrac{dt}{dx}$

$\tt \implies \: - (e ^{ - x} ) dx = dt$

$\tt \implies \: (e ^{ - x} ) dx = - dt$

Now we can write as

$\tt \implies \int {e}^{ - x} cos( {e}^{ - x})dx$

$\tt \implies \: \int - cos(t)dt$

$\tt \implies \: - \int cos(t)dt$

$\tt\implies - sin(t) + c$

Now put the value of t , we get

$\tt\implies - sin({e}^{ - x}) + c$

Answer

$\tt\implies - sin({e}^{ - x}) + c$