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Author: Tinku Tara

I-think-it-will-be-0-pi-4-dx-1-tanx-0-pi-4-dx-1-x-4-1-2-1-2-4-2-1-3-2-4-1-3-4-3-1-3-5-2-4-6-1-4-4-4-

Question Number 86365 by Prithwish Sen 1 last updated on 28/Mar/20 $$\mathrm{I}\:\mathrm{think}\:\mathrm{it}\:\mathrm{will}\:\mathrm{be} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{dx}}{\:\sqrt{\mathrm{1}+\mathrm{tanx}}}\:\approx\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{dx}}{\:\sqrt{\mathrm{1}+\mathrm{x}}}\: \\ $$$$=\frac{\boldsymbol{\pi}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{2}}.\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}\right)^{\mathrm{2}} +\frac{\mathrm{1}.\mathrm{3}}{\mathrm{2}.\mathrm{4}}.\frac{\mathrm{1}}{\mathrm{3}}.\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}\right)^{\mathrm{3}} −\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}}{\mathrm{2}.\mathrm{4}.\mathrm{6}}.\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}\right)^{\mathrm{4}} +…. \\ $$…

Question-151899

Question Number 151899 by liberty last updated on 24/Aug/21 Answered by john_santu last updated on 24/Aug/21 $$\mathrm{A}=\int_{\mathrm{0}} ^{\mathrm{6}} \left(\frac{−\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}+\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx}=\left[−\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{24}}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}\right]_{\mathrm{0}} ^{\mathrm{6}} \\…