1/Sin x'in İntegrali

1/Sin x'in İntegrali Nedir ?

1/Sin x'in İntegrali -ln |csc x+cot x|+c'dir.

$\int \frac{dx}{sinx}=-ln|cscx+cotx|+c$

$\int \frac{dx}{sinx}=ln|cscx-cotx|+c$

$\int \frac{dx}{sinx}=\frac{1}{2}.ln|\frac{1-cosx}{1+cosx}|+c$

$\int \frac{dx}{sinx}=ln|tan\frac{x}{2}|+c$

1/Sin x'in İntegralini Bulma

$\boxed{\frac{1}{sinx}=cscx}$

$\int \frac{1}{sinx}dx=\int cscxdx$

$\int \frac{dx}{sinx}=\int cscxdx$

$\int \frac{dx}{sinx}=\int \frac{(cscx+cotx).cscx}{cscx+cotx}dx$

$\int \frac{dx}{sinx}=\int \frac{cs{c}^{2}x+cscx.cotx}{cscx+cotx}dx$

$cscx+cotx=u$

$d(cscx+cotx)=du$

$(cscx+cotx{)}^{\prime }dx=du$

$(cscx{)}^{\prime }dx+(cotx{)}^{\prime }dx=du$

$\boxed{(cscx{)}^{\prime }=-cscx.cotx}$ $\boxed{(cotx{)}^{\prime }=-cs{c}^{2}x}$

$-cscx.cotxdx-cs{c}^{2}xdx=du$

$(-cscx.cotx-cs{c}^{2}x)dx=du$

$-(cscx.cotx+cs{c}^{2}x)dx=du$

$(cscx.cotx+cs{c}^{2}x)dx=-du$

$\int \frac{dx}{sinx}=\int \frac{-du}{u}$

$\int \frac{dx}{sinx}=-\int \frac{du}{u}$

$\boxed{\int \frac{dx}{x}=ln|x|+c}$

$\int \frac{dx}{sinx}=-ln|u|+c$

$\int \frac{dx}{sinx}=-ln|cscx+cotx|+c$

$-ln|cscx+cotx|=ln|(cscx+cotx{)}^{-1}|$

$\int \frac{dx}{sinx}=ln|(cscx+cotx{)}^{-1}|+c$

$\int \frac{dx}{sinx}=ln|\frac{1}{cscx+cotx}|+c$

$\boxed{cscx=\frac{1}{sinx}}$ $\boxed{cotx=\frac{cosx}{sinx}}$

$\int \frac{dx}{sinx}=ln|\frac{1}{\frac{1}{sinx}+\frac{cosx}{sinx}}|+c$

$\int \frac{dx}{sinx}=ln|\frac{1}{\frac{1+cosx}{sinx}}|+c$

$\int \frac{dx}{sinx}=ln|\frac{sinx}{1+cosx}|+c$

$\int \frac{dx}{sinx}=ln|\frac{sinx.sinx}{sinx.(1+cosx)}|+c$

$\int \frac{dx}{sinx}=ln|\frac{si{n}^{2}x}{sinx.(1+cosx)}|+c$

$\boxed{si{n}^{2}x=1-co{s}^{2}x}$

$\int \frac{dx}{sinx}=ln|\frac{1-co{s}^{2}x}{sinx.(1+cosx)}|+c$

$\int \frac{dx}{sinx}=ln|\frac{(1-cosx).\cancel{(1+cosx)}}{sinx.\cancel{(1+cosx)}}|+c$

$\int \frac{dx}{sinx}=ln|\frac{1-cosx}{sinx}|+c$

$\int \frac{dx}{sinx}=ln|\frac{1}{sinx}-\frac{cosx}{sinx}|+c$

$\int \frac{dx}{sinx}=ln|cscx-cotx|+c$

$cscx-cotx=(\frac{1}{cscx-cotx}{)}^{-1}$

$\int \frac{dx}{sinx}=ln|(\frac{1}{cscx-cotx}{)}^{-1}|+c$

$\int \frac{dx}{sinx}=-ln|\frac{1}{cscx-cotx}|+c$

$\int \frac{dx}{sinx}=-ln|\frac{1}{\frac{1}{sinx}-\frac{cosx}{sinx}}|+c$

$\int \frac{dx}{sinx}=-ln|\frac{1}{\frac{1-cosx}{sinx}}|+c$

$\int \frac{dx}{sinx}=-ln|\frac{sinx}{1-cosx}|+c$

$\boxed{sinx=\sqrt{1-co{s}^{2}x}}$

$\int \frac{dx}{sinx}=-ln|\frac{\sqrt{1-co{s}^{2}x}}{1-cosx}|+c$

$\int \frac{dx}{sinx}=-ln|\frac{\sqrt{1-co{s}^{2}x}}{\sqrt{(1-cosx{)}^2}}|+c$

$\int \frac{dx}{sinx}=-ln|\sqrt{\frac{1-co{s}^{2}x}{(1-cosx{)}^2}}|+c$

$\int \frac{dx}{sinx}=-ln|\sqrt{\frac{\cancel{(1-cosx)}.(1+cosx)}{\cancel{(1-cosx)}.(1-cosx)}}|+c$

$\int \frac{dx}{sinx}=-ln|(\sqrt{\frac{1+cosx}{1-cosx}})|+c$

$\int \frac{dx}{sinx}=ln|(\sqrt{\frac{1+cosx}{1-cosx}}{)}^{-1}|+c$

$\int \frac{dx}{sinx}=ln|\sqrt{(\frac{1+cosx}{1-cosx}{)}^{-1}}|+c$

$\int \frac{dx}{sinx}=ln|\sqrt{\frac{1-cosx}{1+cosx}}|+c$

$\int \frac{dx}{sinx}=ln|(\frac{1-cosx}{1+cosx}{)}^{\frac{1}{2}}|+c$

$\int \frac{dx}{sinx}=\frac{1}{2}.ln|\frac{1-cosx}{1+cosx}|+c$

$\boxed{cos2x=2.co{s}^{2}x-1}$

$\int \frac{dx}{sinx}=\frac{1}{2}.ln|\frac{1-(2.co{s}^2\frac{x}{2}-1)}{1+2.co{s}^2\frac{x}{2}-1}|+c$

$\int \frac{dx}{sinx}=\frac{1}{2}.ln|\frac{1-2.co{s}^2\frac{x}{2}+1}{\cancel{1}+2.co{s}^2\frac{x}{2}-\cancel{1}}|+c$

$\int \frac{dx}{sinx}=\frac{1}{2}.ln|\frac{2-2.co{s}^2\frac{x}{2}}{2.co{s}^2\frac{x}{2}}|+c$

$\int \frac{dx}{sinx}=\frac{1}{2}.ln|\frac{\cancel{2}.(1-co{s}^2\frac{x}{2})}{\cancel{2}.co{s}^2\frac{x}{2}}|+c$

$\int \frac{dx}{sinx}=\frac{1}{2}.ln|\frac{1-co{s}^2\frac{x}{2}}{co{s}^2\frac{x}{2}}|+c$

$\boxed{1-co{s}^{2}x=si{n}^{2}x}$

$\int \frac{dx}{sinx}=\frac{1}{2}.ln|\frac{si{n}^2\frac{x}{2}}{co{s}^2\frac{x}{2}}|+c$

$\int \frac{dx}{sinx}=\frac{1}{2}.ln|(\frac{sin\frac{x}{2}}{cos\frac{x}{2}}{)}^2|+c$

$\int \frac{dx}{sinx}=\cancel{2}.\frac{1}{\cancel{2}}.ln|\frac{sin\frac{x}{2}}{cos\frac{x}{2}}|+c$

$\int \frac{dx}{sinx}=ln|\frac{sin\frac{x}{2}}{cos\frac{x}{2}}|+c$

$\boxed{\frac{sinx}{cosx}=tanx}$

$\int \frac{dx}{sinx}=ln|tan\frac{x}{2}|+c$