Cos x'in İntegrali (Kosinüsün İntegrali)

Cos x'in İntegrali Nedir ?

Cos x'in integrali sin x'tir.

$\int cosxdx=sinx+c$

Cos x'in İntegralini Bulma

1. Yol

Yukarıdaki ABC dik üçgeninde;

$sinx=\frac{u}{1}=u$

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

$\int cosxdx=?$

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

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

$(cosx{)}^{\prime }dx=(\sqrt{1-u^2}{)}^{\prime }du$

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

$\boxed{f(x)=\sqrt{u(x)}\Rightarrow f^{\prime }(x)=\frac{u^{\prime }(x)}{2\sqrt{u(x)}}}$

$-sinxdx=\frac{(1-u^2{)}^{\prime }}{2\sqrt{1-u^2}}du$

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

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

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

$\int cosxdx=u+c$

$\int cosxdx=sinx+c$

2. Yol

$\boxed{sinx=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-...}$

$\boxed{cosx=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-...}$

$\int cosxdx=\int (1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-...)dx$

$\int cosxdx=(x-\frac{x^3}{3.2!}+\frac{x^5}{5.4!}-\frac{x^7}{7.6!}+\frac{x^9}{9.8!}-...)+c$

$\int cosxdx=(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-...)+c$

$c\in R,c=0\Rightarrow \int cosxdx=(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-...)+0$

$\int cosxdx=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-...$

$\int cosxdx=sinx$