Question Number 216776 by MrGaster last updated on 19/Feb/25 Answered by MathematicalUser2357 last updated on 22/Feb/25 $$\mathrm{0}.\mathrm{308425}\:\mathrm{is}\:\mathrm{what}\:\mathrm{I}\:\mathrm{get}. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 216774 by Nadirhashim last updated on 19/Feb/25 $$\:\:\boldsymbol{{find}}\:\int\:\frac{\boldsymbol{{tan}}^{\mathrm{2}} \left(\boldsymbol{{x}}\right)\:}{\mathrm{1}+\boldsymbol{{sec}}^{\mathrm{4}} \left(\boldsymbol{{x}}\right)}\:.\boldsymbol{{dx}}\: \\ $$ Answered by MathematicalUser2357 last updated on 25/Feb/25 $$\frac{\mathrm{1}}{\mathrm{2}}\left\{−\mathrm{2}{x}+\sqrt{\mathrm{1}−{i}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{tan}\left({x}\right)}{\:\sqrt{\mathrm{1}−{i}}}\right)+\sqrt{\mathrm{1}+{i}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{tan}\left({x}\right)}{\:\sqrt{\mathrm{1}+{i}}}\right)\right\}+{C} \\…
Question Number 216772 by Nadirhashim last updated on 19/Feb/25 $$\:\:\boldsymbol{{find}}\:\int\frac{\boldsymbol{{tan}}^{\mathrm{2}} \left(\boldsymbol{{x}}\right)\:}{\mathrm{1}−\boldsymbol{{sec}}^{\mathrm{4}} \left(\boldsymbol{{x}}\right)}\:.\boldsymbol{{dx}}\:\: \\ $$ Answered by MrGaster last updated on 19/Feb/25 $$\:\int\frac{\boldsymbol{{tan}}^{\mathrm{2}} \left(\boldsymbol{{x}}\right)\:}{\mathrm{1}−\boldsymbol{{sec}}^{\mathrm{4}} \left(\boldsymbol{{x}}\right)}\:.\boldsymbol{{dx}}\:=\int\frac{\mathrm{tan}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}}…
Question Number 216754 by Tawa11 last updated on 17/Feb/25 $$\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{x}\:\mathrm{ln}^{\mathrm{2}} \left(\mathrm{x}\right)}{\mathrm{1}\:+\:\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx} \\ $$ Answered by issac last updated on 18/Feb/25 $$\int_{\:\mathrm{0}} ^{\:\mathrm{1}}…
Question Number 216742 by Tawa11 last updated on 17/Feb/25 $$\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{a}\:\:−\:\:\mathrm{1}\right)}{\mathrm{a}}\:\mathrm{da} \\ $$ Answered by sniper237 last updated on 17/Feb/25 $${Not}\:{defined}\:! \\ $$$${But}\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}}…
Question Number 216715 by Nadirhashim last updated on 16/Feb/25 $$\:\:\:\:\boldsymbol{{F}}{ind}\:\int\frac{\boldsymbol{{S}}{in}\left(\frac{\mathrm{5}{x}\:}{\mathrm{2}\:}\right)\:\:}{\boldsymbol{{S}}{in}\left(\frac{{x}\:}{\mathrm{2}\:}\right)\:\:\:\:\:}\:.\boldsymbol{{dx}}\:\:\: \\ $$ Answered by Frix last updated on 16/Feb/25 $$\int\frac{\mathrm{sin}\:\frac{\mathrm{5}{x}}{\mathrm{2}}}{\mathrm{sin}\:\frac{{x}}{\mathrm{2}}}{dx}=\int\left(\mathrm{2cos}\:\mathrm{2}{x}\:+\mathrm{2cos}\:{x}\:+\mathrm{1}\right){dx}= \\ $$$$=\mathrm{sin}\:\mathrm{2}{x}\:+\mathrm{2sin}\:{x}\:+{x}+{C} \\ $$ Answered…
Question Number 216618 by Tawa11 last updated on 12/Feb/25 $$\int_{\:\mathrm{0}} ^{\:\mathrm{2}\pi} \:\sqrt{\mathrm{1}\:\:−\:\:\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\:\:\mathrm{dx} \\ $$$$\mathrm{Is}\:\mathrm{the}\:\mathrm{answer}\:\:\mathrm{0}\:\:\:\mathrm{or}\:\:\:\:\mathrm{4}???? \\ $$ Answered by A5T last updated on 12/Feb/25 $$\int_{\mathrm{0}}…
Question Number 216613 by ajfour last updated on 12/Feb/25 $$\int_{{a}} ^{\:{x}} \frac{\mathrm{1}−{b}\mathrm{ln}\:\frac{{x}}{{a}}}{\:\sqrt{\mathrm{1}−\left(\mathrm{1}−{b}\mathrm{ln}\:\frac{{x}}{{a}}\right)^{\mathrm{2}} }}\:{dx} \\ $$ Commented by ajfour last updated on 12/Feb/25 https://youtu.be/Kq4VqD0azog?si=K_P6MADAgm8CXegJ Terms of…
Question Number 216579 by MrGaster last updated on 11/Feb/25 $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{tan}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)+\mathrm{sin}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)}{\mathrm{tan}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)+\mathrm{cos}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)}{dxdy} \\ $$ Terms…
Question Number 216491 by Jubr last updated on 09/Feb/25 Answered by MrGaster last updated on 09/Feb/25 $$\left(\mathrm{1}\right): \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}\int_{\mathrm{0}} ^{{x}} \left(\mathrm{1}−{t}^{\mathrm{2}} +\frac{{t}^{\mathrm{4}} }{\mathrm{2}!}−\frac{{t}^{\mathrm{6}} }{\mathrm{3}!}+\ldots\right){dt}}{\:\sqrt{\mathrm{1}−{e}^{−{x}^{\mathrm{2}}…