Question

Integration secx.log(secx+tanx)dx

Answer 1

$\boxed{\underline{\textsf{Formulas :}}}$

$1.\:\small{\bold{\int uvdx = u\int vdx - \int\{\frac{du}{dx} \int vdx\}dx}}$

$\textsf{where both u and v are functions of x.}$

$2.\: \small{\bold{\int secx\:dx = log(secx+tanx)+C}}$

$\textsf{where C is integral constant.}$

$3.\: \bold{\frac{d(logx)}{dx}=\frac{1}{x}}$

$\boxed{\underline{\textsf{Solution :}}}$

$\textsf{Now,}\:\bold{\int secx\:log(secx+tanx)\:dx}$

$=\bold{log(secx+tanx)\int secx\:dx}$

$\small{\bold{-\int \{\frac{d}{dx}log(secx+tanx)*\int secx\:dx\}}}$

$=\small{\bold{log(secx+tanx)*log(secx+tanx)}}$

$\tiny{\bold{-\int \{ \frac{\frac{d}{dx}(secx+tanx)}{secx+tanx}*log(secx+tanx)\}dx+2C}}$

$\textsf{where 2C is integral constant.}$

$=\bold{\{log(secx+tanx)\}^{2}}$

$\tiny{\bold{-\int \{\frac{secx\:tanx+sec^{2}x}{secx+tanx}*log(secx+tanx)\}dx+2C}}$

$=\bold{\{log(secx+tanx)\}^{2}}$

$\tiny{\bold{-\int \{\frac{secx(secx+tanx)}{secx+tanx}*log(secx+tanx)\}dx+2C}}$

$=\bold{\{log(secx+tanx)\}^{2}}$

$\bold{-\int secx\:log(secx+tanx)\:dx+2C}$

$\to \tiny{\bold{2\int secx\:log(secx+tanx)\:dx = \{log(secx+tanx)\}^{2}+2C}}$

$\to \boxed{\tiny{\bold{\int secx\:log(secx+tanx)\:dx=\frac{1}{2}\{log(secx+tanx)\}^{2}+C}}}$

$\textsf{which is the required integral.}$