Question Number 120254 by bramlexs22 last updated on 30/Oct/20 $$\:\int\:\frac{{dx}}{\mathrm{1}+\mathrm{cos}\theta.\mathrm{cos}\:{x}\:}\:? \\ $$ Answered by TANMAY PANACEA last updated on 30/Oct/20 $$\int\frac{{dx}}{{cos}\theta\left({sec}\theta+{cosx}\right)} \\ $$$$\frac{\mathrm{1}}{{cos}\theta}\int\frac{{dx}}{{sec}\theta+\frac{\mathrm{1}−{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\mathrm{1}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}}…
Question Number 54701 by Meritguide1234 last updated on 09/Feb/19 Answered by behi83417@gmail.com last updated on 09/Feb/19 $${tg}^{−\mathrm{1}} \frac{{x}}{{x}^{\mathrm{2}} −\mathrm{1}}={a} \\ $$$${sin}^{\mathrm{2}} {a}=\frac{\mathrm{1}}{\mathrm{1}+{cot}^{\mathrm{2}} {a}}=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} +{x}^{−\mathrm{2}} −\mathrm{2}}=\frac{{x}^{\mathrm{2}}…
Question Number 54699 by rahul 19 last updated on 09/Feb/19 $$\int\:{e}^{\mathrm{2}{x}} \:\left(\frac{\mathrm{1}}{{x}}\:−\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }\right){dx} \\ $$$$\int{e}^{\mathrm{2}{x}} \:\left(\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{1}+\mathrm{cos}\:\mathrm{2}{x}}\right){dx}. \\ $$$${Solve}\:{above}\:{Questions}\:{by}\:{using}\:{the} \\ $$$${formulae}\:: \\ $$$$\int{e}^{{kx}} \left\{{f}\left({kx}\right)+{f}\:'\left({kx}\right)\right\}{dx}=\:{e}^{{kx}} \:{f}\left({kx}\right)+{c}. \\…
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Question Number 54616 by cesar.marval.larez@gmail.com last updated on 07/Feb/19 Answered by tanmay.chaudhury50@gmail.com last updated on 08/Feb/19 $${a}={tanx}\:\:{da}={sec}^{\mathrm{2}} {xdx} \\ $$$$\int\frac{{a}^{\mathrm{2}} {da}}{\:\sqrt{\mathrm{2}+\mathrm{1}+{a}^{\mathrm{2}} }} \\ $$$$\int\frac{{a}^{\mathrm{2}} +\mathrm{3}−\mathrm{3}}{\:\sqrt{{a}^{\mathrm{2}}…
Question Number 185677 by Mingma last updated on 25/Jan/23 Answered by Frix last updated on 25/Jan/23 $$\frac{{df}}{{dx}}=\mathrm{e}^{\mid{x}\mid} −\mathrm{e}^{\mid{x}−\mathrm{2}\mid} =\mathrm{0}\:\Rightarrow\:{x}=\mathrm{1} \\ $$$${f}\left(\mathrm{1}\right)=\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}\mathrm{e}^{\mid\mathrm{1}−{t}\mid} {dt}=\mathrm{2}\underset{\mathrm{0}} {\overset{\mathrm{1}}…
Question Number 54586 by arvinddayama00@gmail.com last updated on 07/Feb/19 $$\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:+\:\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }\:=\:{a}\left({x}−{y}\right) \\ $$$${prove}\:{that} \\ $$$$\frac{{dy}}{{dx}}=\sqrt{\frac{\mathrm{1}−{y}^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated…
Question Number 54585 by peter frank last updated on 07/Feb/19 $${show}\:{that}\:\int_{\mathrm{0}} ^{\infty} \frac{{x}}{\mathrm{1}+{x}^{\mathrm{6}} }{dx}=\frac{\pi}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$$$ \\ $$ Commented by maxmathsup by imad last updated…
Question Number 54573 by Abdo msup. last updated on 07/Feb/19 $${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{cos}\left({e}^{−{x}^{\mathrm{2}} } \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\: \\ $$ Commented by maxmathsup by imad last updated…