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Question Number 106677 by mohammad17 last updated on 06/Aug/20

Answered by Dwaipayan Shikari last updated on 06/Aug/20

$$5){\int }_1^4(1-u)\sqrt{u}du$$$${\int }_1^4\sqrt{u}-u^{\frac{3}{2}}du$$$${[\frac{2}{3}{u}^{\frac{3}{2}}-\frac{2}{5}{u}^{\frac{5}{2}}]}_1^4=\frac{16}{3}-\frac{2}{3}-\frac{64}{5}+\frac{2}{5}=\frac{14}{3}-\frac{62}{5}=\frac{-116}{15}$$

Commented by mohammad17 last updated on 06/Aug/20

$$sircanyoureppedagaintheresult$$

Commented by Her_Majesty last updated on 06/Aug/20

$$resultis-\frac{116}{15}$$

Commented by mohammad17 last updated on 06/Aug/20

$$thankyousir$$

Answered by Dwaipayan Shikari last updated on 06/Aug/20

$$4)\int \frac{tanxsecx}{\sqrt{1+secx}}dx$$$$=\int \frac{2tdt}{t}(1+secx=t^2,secxtanx=2t\frac{dt}{dx}$$$$=2t+C=2\sqrt{1+secx}+C$$

Answered by som(math1967) last updated on 06/Aug/20

$$4)\int \frac{\mathrm{secxtanx}}{\sqrt{1+\mathrm{secx}}}\mathrm{dx}$$$$\mathrm{let}1+\mathrm{secx}={\mathrm{z}}^2$$$$\therefore \mathrm{secxtanxdx}=2\mathrm{zdz}$$$$\therefore \int \frac{2\mathrm{zdz}}{\mathrm{z}}$$$$=2\mathrm{z}+\mathrm{C}$$$$=2\sqrt{1+\mathrm{secx}}+\mathrm{C}$$

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