Question Number 14479 by tawa tawa last updated on 01/Jun/17
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 01/Jun/17
$${ln}\left(\mathrm{1}+{tg}^{\mathrm{2}} {x}\right)={ln}\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {x}}=−{lncos}^{\mathrm{2}} {x}=−\mathrm{2}{lncosx}. \\ $$$${a}^{\frac{\mathrm{3}}{\mathrm{2}}{log}_{\sqrt{{a}}} ^{\left({secx}\right)} } ={a}^{\frac{\mathrm{3}}{\mathrm{2}}\left(−\mathrm{2}{log}_{{a}} {cosx}\right)} ={a}^{{log}_{\boldsymbol{{a}}} ^{\frac{\mathrm{1}}{\boldsymbol{{cos}}^{\mathrm{3}} \boldsymbol{{x}}}} } =\frac{\mathrm{1}}{\boldsymbol{{cos}}^{\mathrm{3}} \boldsymbol{{x}}} \\ $$$$\boldsymbol{{I}}=\int−\frac{\mathrm{2}{lncosx}}{{cos}^{\mathrm{3}} {x}}{dx} \\ $$$${u}={lncosx}\Rightarrow{du}=\frac{−{sinx}}{{cosx}}{dx} \\ $$$$\Rightarrow{I}=\int\frac{\mathrm{2}{sinx}}{{cos}^{\mathrm{4}} {x}}{dx}=\frac{−\mathrm{2}}{\mathrm{5}{cos}^{\mathrm{5}} {x}}+\boldsymbol{{C}}\:.\blacksquare \\ $$
Commented by tawa tawa last updated on 01/Jun/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$