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Question Number 121548 by benjo_mathlover last updated on 09/Nov/20

$$\underset{0}{\stackrel{\pi /4}{\int }}\tan {}^6\mathrm{x}\sec \mathrm{x}\mathrm{dx}=?$$

Answered by MJS_new last updated on 09/Nov/20

$$\int {\tan }^{6}x\sec xdx=\int \frac{{\sin }^{6}x}{{\cos }^{7}x}dx=$$$$[t=\frac{1}{\sin x}\to dx=-\frac{{\sin }^{2}x}{\cos x}]$$$$=-\int \frac{dt}{{(t^2-1)}^4}=$$$$\left[{\mathrm{Ostrogradski}}^{\prime }\mathrm{s}\mathrm{Method}\right]$$$$=\frac{t(15t^4-40t^2+33)}{48{(t^2-1)}^3}+\frac{5}{16}\int \frac{dt}{t^2-1}=$$$$=\frac{t(15t^4-40t^2+33)}{48{(t^2-1)}^3}+\frac{5}{32}\ln \mid \frac{t-1}{t+1}\mid$$$$\mathrm{t}\mathrm{he}\mathrm{borders}\mathrm{are}x\in [0;\frac{\pi }{4}]\iff t\in ]+\mathrm{\infty };\sqrt{2}]$$$$\Rightarrow \mathrm{answer}\mathrm{is}$$$$-{[\frac{t(15t^4-40t^2+33)}{48{(t^2-1)}^3}+\frac{5}{32}\ln \mid \frac{t-1}{t+1}\mid ]}_{\sqrt{2}}^{+\mathrm{\infty }}=$$$$=\frac{13\sqrt{2}}{48}+\frac{5}{16}\ln (-1+\sqrt{2})$$

Commented by liberty last updated on 09/Nov/20

$$\mathrm{waw}...\mathrm{Your}\mathrm{method}\mathrm{good}\mathrm{sir}$$

Commented by Fareed last updated on 09/Nov/20

$$$$$$$$$$\mathrm{please}\mathrm{solve}\mathrm{it}\mathrm{too}$$$$\mathrm{p}(\mathrm{x},\mathrm{y})={2\mathrm{x}}^3\mathrm{y}+{8\mathrm{xy}}^3$$$$\mathrm{find}\mathrm{the}({\mathrm{a}}_4=?)$$

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