∫tan^(1/5) xdx


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Question Number 99485 by Dwaipayan Shikari last updated on 21/Jun/20

$\int {t a n}^{\frac{1}{5}} x d x$
	Answered by MJS last updated on 21/Jun/20

	$t = \tan^{2 / 5} x$ $\int \tan^{1 / 5} x d x =$ $[t = \tan^{2 / 5} x \to d x = \frac{5}{2} \tan^{3 / 5} x \cos^{2} x dt]$ $= \frac{5}{2} \int \frac{t^{2}}{t^{5} + 1} d t$ $the rest is decomposing$ $t^{5} + 1 = (t + 1) (t^{2} - \frac{1 - \sqrt{5}}{2} t + 1) (t^{2} - \frac{1 + \sqrt{5}}{2} t + 1)$
	Commented by PRITHWISH SEN 2 last updated on 21/Jun/20
	$the decomosition −((2(√(10+2(√5))) (1+t^5 ))/(5{2t^2 +2t^3 −(√(10−2(√5))) t^(5/2) })) +(((√(10−2(√5)))(1+t^5 ))/(20{2t^2 +2t^3 −(√(10+2(√5))) t^(5/2) )) +((2(√(10+2(√5)))(1+t^5 ))/({2t^2 +2t^3 +(√(10−2(√5))) t^(5/2) })) + (((√(10−2(√5)))(1+t^5 ))/(20{2t^2 +2t^3 +(√(10+2(√5))) t^(5/2) })) + (((1+t^5 ))/(2(1+t)t^(3/2) )) −(((1+(√5))(1+t^5 )^2 )/(40{t+(√((5−(√5))/2)) (√t) +1})) − (((1+(√5))(1+t^5 )^2 )/(40{t−(√((5−(√5))/2)) (√t)+1})) + ((((√5)−1)(1+t^5 )^2 )/(40{t+(√((5+(√5))/2)) (√t)+1})) {(2/t^(3/2) ) +(√((5+(√5))/2)) .(1/t^2 )} + ((((√5)−1)(1+t^5 )^2 )/(40{t−(√((5+(√5))/2)) (√t)+1})) {(2/t^(3/2) ) −(√((5+(√5))/2)) .(1/t^2 )} I did not check it . Please check.$
	$the decomosition$ $- \frac{2 \sqrt{10 + 2 \sqrt{5}} (1 + t^{5})}{5 {{2 t}^{2} + {2 t}^{3} - \sqrt{10 - 2 \sqrt{5}} t^{\frac{5}{2}}}} + \frac{\sqrt{10 - 2 \sqrt{5}} (1 + t^{5})}{20 {{2 t}^{2} + {2 t}^{3} - \sqrt{10 + 2 \sqrt{5}} t^{\frac{5}{2}}}$ $+ \frac{2 \sqrt{10 + 2 \sqrt{5}} (1 + t^{5})}{{{2 t}^{2} + {2 t}^{3} + \sqrt{10 - 2 \sqrt{5}} t^{\frac{5}{2}}}} +$ $\frac{\sqrt{10 - 2 \sqrt{5}} (1 + t^{5})}{20 {{2 t}^{2} + {2 t}^{3} + \sqrt{10 + 2 \sqrt{5}} t^{\frac{5}{2}}}} +$ $\frac{(1 + t^{5})}{2 (1 + t) t^{\frac{3}{2}}} - \frac{(1 + \sqrt{5}) {(1 + t^{5})}^{2}}{40 {t + \sqrt{\frac{5 - \sqrt{5}}{2}} \sqrt{t} + 1}}$ $- \frac{(1 + \sqrt{5}) {(1 + t^{5})}^{2}}{40 {t - \sqrt{\frac{5 - \sqrt{5}}{2}} \sqrt{t} + 1}} +$ $\frac{(\sqrt{5} - 1) {(1 + t^{5})}^{2}}{40 {t + \sqrt{\frac{5 + \sqrt{5}}{2}} \sqrt{t} + 1}} {\frac{2}{t^{\frac{3}{2}}} + \sqrt{\frac{5 + \sqrt{5}}{2}} . \frac{1}{t^{2}}}$ $+ \frac{(\sqrt{5} - 1) {(1 + t^{5})}^{2}}{40 {t - \sqrt{\frac{5 + \sqrt{5}}{2}} \sqrt{t} + 1}} {\frac{2}{t^{\frac{3}{2}}} - \sqrt{\frac{5 + \sqrt{5}}{2}} . \frac{1}{t^{2}}}$ $I did not check it . Please check .$
	Commented by PRITHWISH SEN 2 last updated on 21/Jun/20

	Commented by PRITHWISH SEN 2 last updated on 21/Jun/20

	$from wolfram alpha$
	Commented by MJS last updated on 21/Jun/20

	${that}^{'} s exactly where my path leads to, no$ $machine logic needed . {it}^{'} s just horrible$ $constants once again$
	Answered by maths mind last updated on 22/Jun/20

	$u = t a n (x) \Leftrightarrow \int \frac{u^{\frac{1}{5}}}{1 + u^{2}} d u$ $= \sum_{k ⩾ 0} \int u^{\frac{1}{5}} {(- u^{2})}^{k} d u$ $= \sum_{k ⩾ 0} \int {(- 1)}^{k} u^{2 k + \frac{1}{5}} d u$ $= \sum_{k ⩾ 0} \frac{{(- 1)}^{k} u^{2 k + \frac{6}{5}}}{(2 k + \frac{6}{5})} = u^{\frac{6}{5}} Σ \frac{{(- u^{2})}^{k}}{2 (k + \frac{3}{5})} + c$ $= \frac{5}{6} u^{\frac{6}{5}} \sum_{k ⩾ 0} \frac{\frac{3}{5}}{k + \frac{3}{5}} {(- u^{2})}^{k} + c$ $= \frac{5}{6} u^{\frac{6}{5}} (1 + \sum_{k ⩾ 1} \frac{\underset{j = 0}{\prod^{k - 1}} (\frac{3}{5} + j) . \underset{j = 0}{\prod^{k - 1}} (1 + j)}{\underset{j = 0}{\prod^{k - 1}} (\frac{8}{5} + j)} . \frac{{(- u^{2})}^{k}}{k!}) + c$ $= \frac{5}{6} u^{\frac{6}{5}}_{2} F_{1} (\frac{3}{5}, 1; \frac{8}{5}; - u^{2}) + c$ $\Leftrightarrow \int {t g}^{\frac{1}{5}} (x) d x = \frac{5}{6} {t g}^{\frac{6}{5}} (x)_{2} F_{1} (\frac{3}{5}, 1; \frac{8}{5}; - {t g}^{2} (x)) + c$
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