Question Number 201221 by Calculusboy last updated on 02/Dec/23 Answered by MM42 last updated on 02/Dec/23 $$\bigstar\bigstar\bigstar\:\:{tan}^{−\mathrm{1}} {a}−{tan}^{−\mathrm{1}} {b}={tan}^{−\mathrm{1}} \left(\frac{{a}−{b}}{\mathrm{1}+{ab}}\right) \\ $$$$\Rightarrow{tan}^{−\mathrm{1}} \left(\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}\right)−{tan}^{−\mathrm{1}} \left(\frac{{x}}{{x}+\mathrm{2}}\right)={tan}^{−\mathrm{1}} \left(\frac{{x}+\mathrm{2}}{\mathrm{2}{x}^{\mathrm{2}}…
Question Number 201222 by Calculusboy last updated on 02/Dec/23 $$\:\int\:\left(\boldsymbol{{x}}^{\mathrm{6}} +\boldsymbol{{x}}^{\mathrm{9}} \right)^{\frac{\mathrm{1}}{\mathrm{6}}} \boldsymbol{{dx}} \\ $$ Answered by MathematicalUser2357 last updated on 04/Jan/24 $$\mathrm{A}\:\mathrm{function}\:\mathrm{that}\:\mathrm{contains}\:_{\mathrm{2}} {F}_{\mathrm{1}} \\…
Question Number 201223 by Calculusboy last updated on 02/Dec/23 $$\int\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\boldsymbol{{x}}}}\:\boldsymbol{{dx}} \\ $$ Answered by witcher3 last updated on 02/Dec/23 $$=\int\mathrm{2}\sqrt{\mathrm{1}+\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}}}.\frac{\mathrm{dx}}{\mathrm{2}\:\sqrt{\mathrm{1}+\mathrm{x}}} \\ $$$$=\mathrm{2}\int\sqrt{\mathrm{t}^{\mathrm{2}} +\mathrm{t}}\mathrm{dt} \\ $$$$\Leftrightarrow\int\mathrm{2}\sqrt{\left(\mathrm{t}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}}…
Question Number 201172 by Calculusboy last updated on 01/Dec/23 Answered by Sutrisno last updated on 01/Dec/23 $$=\int\frac{\mathrm{2}{e}^{\mathrm{2}{x}} −{e}^{{x}} }{\:\sqrt{\mathrm{3}\left({e}^{\mathrm{2}{x}} −\mathrm{2}{e}^{{x}} −\frac{\mathrm{1}}{\mathrm{3}}\right)}}{dx} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\int\frac{\mathrm{2}{e}^{\mathrm{2}{x}} −{e}^{{x}} }{\:\sqrt{\left({e}^{{x}}…
Question Number 201200 by Calculusboy last updated on 01/Dec/23 Commented by Rasheed.Sindhi last updated on 02/Dec/23 $$\left(\:\left({x}−\mathrm{1}\right)^{{x}−\mathrm{1}} \:\right)^{\mathrm{1}/\mathrm{3}} =\left({x}−\mathrm{1}\right)^{\left({x}−\mathrm{1}\right)^{\mathrm{1}/\mathrm{3}} } \\ $$ Commented by Rasheed.Sindhi…
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Question Number 201166 by mnjuly1970 last updated on 01/Dec/23 Answered by mr W last updated on 01/Dec/23 $$\frac{{ah}}{\mathrm{2}}=\mathrm{4} \\ $$$$\Rightarrow{h}=\frac{\mathrm{2}×\mathrm{4}}{\:\sqrt{\mathrm{2}}}=\mathrm{4}\sqrt{\mathrm{2}} \\ $$$${h}×\mathrm{cot}\:{B}+{h}×\mathrm{cot}\:{C}={a} \\ $$$$\Rightarrow\mathrm{cot}\:{B}+\mathrm{cot}\:{C}=\frac{{a}}{{h}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}\sqrt{\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{4}} \\…
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