1/Sin x'in İntegrali

1/Sin x'in İntegrali Nedir ?

$\int \frac{dx}{sinx}=-ln|cscx+cotx|+c=ln|cscx-cotx|+c=ln|tan\frac{x}{2}|+c=ln|\sqrt{\frac{cosx-1}{cosx+1}}|+c$

1/Sin x'in İntegralini Bulma

1. Yol

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

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

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

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

$cscx+cotx=u$

$d(cscx+cotx)=du$

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

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

$\boxed{(cscx{)}^{\prime }=-cscx.cotx}$ $\boxed{(cotx{)}^{\prime }=-(1+co{t}^{2}x)=-\frac{1}{si{n}^{2}x}=-cs{c}^{2}x}$

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

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

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

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

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

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

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

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

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

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

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

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

$\int \frac{dx}{sinx}=ln|\frac{cscx-cotx}{cs{c}^{2}x-co{t}^{2}x}|+c$

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

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

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

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

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

$\int \frac{dx}{sinx}=ln|\frac{cscx-cotx}{\frac{\cancel{si{n}^{2}x}}{\cancel{si{n}^{2}x}}}|+c$

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

2. Yol

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

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

$\int \frac{dx}{sinx}=\int \frac{sinx}{si{n}^{2}x}dx$

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

$\int \frac{dx}{sinx}=\int \frac{sinx}{1-co{s}^{2}x}dx$

$cosx=u$

$d(cosx)=du$

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

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

$-sinxdx=du$

$sinxdx=-du$

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

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

$\int \frac{dx}{sinx}=\int \frac{du}{(u-1).(u+1)}$

$\frac{1}{(u-1).(u+1)}=\frac{A}{u-1}+\frac{B}{u+1}$

$\frac{1}{(u-1).(u+1)}=\frac{A.(u+1)+B.(u-1)}{(u-1).(u+1)}$

$\frac{1}{\cancel{(u-1).(u+1)}}=\frac{Au+A+Bu-B}{\cancel{(u-1).(u+1)}}$

$0.u+1=(A+B).u+A-B$

$A+B=0$

$A-B=1$

$A+\cancel{B}+A-\cancel{B}=0+1$

$2A=1$

$A=\frac{1}{2}$

$\frac{1}{2}+B=0$

$B=0-\frac{1}{2}=-\frac{1}{2}$

$\int \frac{dx}{sinx}=\int \frac{1/2}{u-1}du+\int \frac{-1/2}{u+1}du$

$\int \frac{dx}{sinx}=\int \frac{1/2}{u-1}du-\int \frac{1/2}{u+1}du$

$\int \frac{dx}{sinx}=\frac{1}{2}\int \frac{du}{u-1}-\frac{1}{2}\int \frac{du}{u+1}$

$\int \frac{dx}{sinx}=\frac{1}{2}(\int \frac{du}{u-1}-\int \frac{du}{u+1})$

$\boxed{\int \frac{dx}{x\pm a}=ln|x\pm a|}$

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

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

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

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

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

3. Yol

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

$\boxed{sin2x=2.sinx.cosx}$

$\int \frac{1}{sinx}dx=\int \frac{si{n}^2\frac{x}{2}+co{s}^2\frac{x}{2}}{2.sin\frac{x}{2}.cos\frac{x}{2}}dx$

$\int \frac{dx}{sinx}=\int (\frac{si{n}^2\frac{x}{2}}{2.sin\frac{x}{2}.cos\frac{x}{2}}+\frac{co{s}^2\frac{x}{2}}{2.sin\frac{x}{2}.cos\frac{x}{2}})dx$

$\int \frac{dx}{sinx}=\int \frac{si{n}^2\frac{x}{2}}{2.sin\frac{x}{2}.cos\frac{x}{2}}dx+\int \frac{co{s}^2\frac{x}{2}}{2.sin\frac{x}{2}.cos\frac{x}{2}}dx$

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

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

$\int \frac{dx}{sinx}=\frac{1}{2}\int \frac{sin\frac{x}{2}}{cos\frac{x}{2}}dx+\frac{1}{2}\int \frac{cos\frac{x}{2}}{sin\frac{x}{2}}dx$

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

$\int \frac{dx}{sinx}=\frac{1}{2}\int \tan \frac{x}{2}dx+\frac{1}{2}\int \cot \frac{x}{2}dx$

$\boxed{\int tanaxdx=\frac{1}{a}ln|\frac{1}{cosax}|+c}$

$\boxed{\int cotaxdx=\frac{1}{a}ln|sinax|+c}$

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

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

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

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

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