Question

Find integrate of (sin^2x-cos^2x)/(sinx×cosx) dx.​​

Answer 1

To Integrate :

$\star \: \int \: \frac{ {sin}^{2}x - {cos}^{2} x }{sinx \: cosx} dx$

$\implies \: \int \: \frac{ { sin}^{2}x }{sinx \: cosx} dx - \int \: \frac{ { cos}^{2} x}{ sinx \: cosx \: } dx \\ \\ \implies \int \: \frac{ { \cancel{sin}}^{2}x }{ \cancel{sinx}cosx} dx - \int \: \frac{ { \cancel{cos}}^{2}x }{ sinx \cancel{cosx}} dx \\ \\ \implies \: \int \: \frac{sinx}{cosx} dx - \int \: \frac{cosx}{sinx} dx \\ \\ \implies \int \: tanx \: dx - \int \: cotx \: dx \\ \\ \implies \: ln(secx) - ln(sinx) + c \\ \\ \implies \: ln( \frac{secx}{sinx} ) + c$

Formula :

$\star \boxed{\bold{ \int \: tanx \: dx = ln(secx) + c}} \\ \\ \star \boxed{ \bold{ \int \: cotx \: dx = ln(sinx) + c}}$

Answer 2

Step-by-step explanation:

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