Sin x'in İntegrali

Sin x'in İntegrali Nedir ?

Sin x'in integrali -cos x+c'dir.

$\int sinxdx=-cosx+c$

Sin x'in İntegralini Bulma

$\int sinxdx=?$

$sinx=u$

$d(sinx)=du$

$(sinx{)}^{\prime }dx=du$

$\boxed{(sinx{)}^{\prime }=cosx}$

$cosxdx=du$

$dx=\frac{du}{cosx}$

$sinx=u$

$cosx=\sqrt{1-u^2}$

$dx=\frac{du}{\sqrt{1-u^2}}$

$\int sinxdx=\int u.\frac{du}{\sqrt{1-u^2}}$

$\int sinxdx=\int \frac{udu}{\sqrt{1-u^2}}$

$\sqrt{1-u^2}=v$

$(\sqrt{1-u^2}{)}^2=v^2$

$\sqrt{(1-u^2{)}^2}=v^2$

$1-u^2=v^2$

$d(1-u^2)=d(v^2)$

$(1-u^2{)}^{\prime }du=(v^2{)}^{\prime }dv$

$-\cancel{2}udu=\cancel{2}vdv$

$-udu=vdv$

$udu=-vdv$

$\int sinxdx=\int \frac{-\cancel{v}dv}{\cancel{v}}$

$\int sinxdx=\int -dv$

$\int sinxdx=-v+c$

$\int sinxdx=-\sqrt{1-u^2}+c$

$\int sinxdx=-cosx+c$