Question

Integration of sin x [log cos x] dx

Answer 1

Answer:

_cos x log sin x

here is your answer

Answer 2

Answer:

The Integration for the given function will be:

$-cosx\ log(cosx)+cosx+c$

Step-by-step explanation:

Given integration is:

$\displaystyle\int sinx[log\ cosx]\ dx$

Here we are applying the formula of integration by parts method that has been shown below:

$\displaystyle\int (uv)dx=u\displaystyle\int v\ dx-\displaystyle\int [\dfrac{d}{dx}(u)\displaystyle\int v\ dx]dx$

Here:

$u=log(cosx)\\v=sinx$

Hence integrating:

$=log(cosx)\displaystyle\int sinx\ dx-\displaystyle\int [\dfrac{d}{dx}[log\ cosx]\displaystyle\int sinx\ dx]dx$

Hence we know the formula of integration:

$\displaystyle\int sinx\ dx=-cosx+c\\ \displaystyle\int cosx\ dx=sinx+c$

Therefore:

$=log(cosx)\cdot (-cosx)-\displaystyle\int [\dfrac{1}{cosx}(-sinx)\cdot (-cosx)]dx\\=-cosx\ log(cosx)-\displaystyle\int sinx\ dx\\=-cosx\ log(cosx)+cosx+c$

Here c is the integration constant.