★ Formula Applied :
[tex]\begin{gathered}\sf \bullet\ \; cos^2x-sin^2x=cos2x\\\\\to\ \sf \pink{sin^2x-cos^2x=-cos2x}\end{gathered}[/tex]
[tex]\begin{gathered}\bullet\ \; \sf sin2x=2.sinx.cosx\\\\\to \sf \blue{sinx.cosx=\dfrac{sin2x}{2}}\end{gathered}[/tex]
[tex]\displaystyle \bullet\ \; \sf \int \dfrac{dx}{x}[/tex]
[tex]\bullet\ \; \sf \ln (ab)=\ln a+\ln[/tex]
★ Explanation :
[tex]\begin{gathered}\displaystyle \sf \int \dfrac{sin^2x-cos^2x}{sinx.cosx}dx\\\\\to \sf \int \dfrac{-cos2x}{\frac{sin2x}{2}}dx\\\\\to \sf \int \dfrac{-2cos2x}{sin2x}dx\end{gathered}[/tex]
Lets use substitution method ,
Let , u = sin2x
⇒ du = 2.cos2x.dx
[tex]\begin{gathered}\to \displaystyle \sf \int \dfrac{-du}{u}\\\\\to \sf -\int \dfrac{du}{u}\\\\\to \sf -ln|u|\\\\\end{gathered}[/tex]
[tex]\to \sf \red{-ln|sin2x|+c}[/tex]
[tex]\to \sf - \ln |2sinx.cosx|+c[/tex]
[tex]\to \sf - \ln |sinx.cosx|+(\ln 2+c)[/tex]
[tex]\leadsto - \sf \red{\ln |sinx.cosx|+c}\ \; \bigstar[/tex]
★ Alternate Method :
[tex]\displaystyle \sf \int \dfrac{sin^2x-cos^2x}{sinx.cosx}[/tex]
[tex]\displaystyle \to \sf \int \left( \dfrac{sin^2x}{sinx.cosx}-\dfrac{cos^2x}{sinx.cosx}\right)dx[/tex]
[tex]\begin{gathered}\displaystyle \to \sf \int \left( \dfrac{sinx}{cosx} -\dfrac{cosx}{sinx} \right)dx\\\\\to\ \sf \int (tanx-cotx)dx\\\\\to \sf \int tanx.dx-\int cotx.dx\\\\\to \sf ln|secx|-ln|sinx|+c\end{gathered}[/tex]
[tex]\to \sf ln\left| \dfrac{secx}{sinx}\right|+c[/tex]
[tex]\begin{gathered}\to \sf ln\left| \dfrac{1}{sinx.cosx} \right|+c\\\\\end{gathered}[/tex]]
[tex]\leadsto \sf \pink{-ln|sinx.cosx|+c}\ \; \bigstar[/tex]
