Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 136043 by JulioCesar last updated on 18/Mar/21

Answered by rs4089 last updated on 18/Mar/21

$$\int {sec}^{8}x.dx$$$$\int {(1+{tan}^{2}x)}^3{sec}^{2}x.dx$$$$tanx=t\Rightarrow {sec}^{2}x.dx=dt$$$$\int {(1+t^2)}^{3}dt$$$$\int (1+t^6+3t^2+3t^4)dt$$$$t+\frac{t^7}{7}+t^3+3\frac{t^5}{5}+C$$$$tanx+\frac{{tan}^{7}x}{7}+{tan}^{3}x+3\frac{{tan}^{5}x}{5}+C$$

Answered by Olaf last updated on 18/Mar/21

$$\mathrm{I}=\int \frac{dx}{{\cos }^{8}x}$$$$\mathrm{I}=\int (\frac{1}{{\cos }^{6}x}+\frac{{\sin }^{2}x}{{\cos }^{8}x})dx$$$$\mathrm{I}=\int (\frac{1}{{\cos }^{4}x}+\frac{{\sin }^{2}x}{{\cos }^{6}x}+\frac{{\sin }^{2}x}{{\cos }^{8}x})dx$$$$\mathrm{I}=\int (\frac{1}{{\cos }^{2}x}+\frac{{\sin }^{2}x}{{\cos }^{4}x}+\frac{{\sin }^{2}x}{{\cos }^{6}x}+\frac{{\sin }^{2}x}{{\cos }^{8}x})dx$$$$\mathrm{I}=\int (1+\frac{{\sin }^{2}x}{{\cos }^{2}x}+\frac{{\sin }^{2}x}{{\cos }^{4}x}+\frac{{\sin }^{2}x}{{\cos }^{6}x}+\frac{{\sin }^{2}x}{{\cos }^{8}x})dx$$$$\mathrm{I}=x+\int \sin x(\frac{\sin x}{{\cos }^{2}x}+\frac{\sin x}{{\cos }^{4}x}+\frac{\sin x}{{\cos }^{6}x}+\frac{\sin x}{{\cos }^{8}x})dx$$$$\mathrm{I}=x+\sin x(\frac{1}{\cos x}+\frac{1}{{3\cos }^{3}x}+\frac{1}{{5\cos }^{5}x}+\frac{1}{{\cos }^{7}x})$$$$-\int \cos x(\frac{1}{\cos x}+\frac{1}{{3\cos }^{3}x}+\frac{1}{{5\cos }^{5}x}+\frac{1}{{7\cos }^{7}x})dx$$$$\mathrm{I}=\sin x(\frac{1}{\cos x}+\frac{1}{{3\cos }^{3}x}+\frac{1}{{5\cos }^{5}x}+\frac{1}{{\cos }^{7}x})$$$$-\int (\frac{1}{{3\cos }^{2}x}+\frac{1}{{5\cos }^{4}x}+\frac{1}{{7\cos }^{6}x})dx$$$$...etc$$$$\mathrm{I}=\frac{\sin x}{35}(\frac{16}{\cos x}+\frac{8}{{\cos }^{3}x}+\frac{6}{{\cos }^{5}x}+\frac{5}{{\cos }^{7}x})+\mathrm{C}$$$$\mathrm{I}=\frac{\sin x}{35}(16\sec x+{8\sec }^{3}x+{6\cos }^{5}x+{5\sec }^7)+\mathrm{C}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com