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Question Number 120059 by A8;15: last updated on 29/Oct/20

	Answered by Lordose last updated on 29/Oct/20
	$Firstly, compute the indefinite integral I = ∫xsec(2x)dx = (1/4)∫asec(a)da {a=2x} I = (1/4)(alntan((a/2)+(π/4)) − ∫lntan((a/2)+(π/4))da) {IBP} set (a/2) + (π/4) = y ⇒ da=2dy I = (1/4)(alntan(y) − 2∫lntan(y)dy) Ω = ∫lntan(y)dy {tan(y) = ((1−e^(−2iy) )/(1+e^(−2iy) ))} Ω = ∫ln(1−e^(−2iy) )dy − ∫ln(1+e^(−2iy) )dy set w=e^(−2iy ) ⇒dw=((2w)/i)dy Ω = (i/2)∫ ((ln(1−w))/w)dw − (i/2)∫ ((ln(1+w))/w)dw Ω = −(i/2)Li_2 (w) − (i/2)(−Li_2 (−w)) + C Ω = (i/2)(Li_2 (−e^(−2iy) ) − Li_2 (e^(−2iy) )) I=(1/4)(alntan(y) − i(Li_2 (−e^(−2iy) )−Li_2 (e^(−2iy) ))) I = (1/2)xlntan(x+(π/4)) − (i/4)(Li_2 (−e^(−i(2x+(π/2))) ) − Li_2 (e^(−i(2x+(π/2))) )) + C ∫_0 ^( 1) xsec(2x)dx =( (1/2)lntan(((4+π)/4)) − (i/4)Li_2 (−e^(−i(((4+π)/2))) ) + (i/4)Li_2 (e^(−i(((4+π)/2))) )) + (i/4)(Li_2 (−e^(−((iπ)/2)) )− Li_2 (e^(−((iπ)/2)) )) ∫_0 ^( 1) xsec(2x)dx = ∞ The integral diverges.$
	$Firstly, compute the indefinite integral$ $I = \int xsec (2 x) dx = \frac{1}{4} \int asec (a) da {a = 2 x}$ $I = \frac{1}{4} (alntan (\frac{a}{2} + \frac{π}{4}) - \int lntan (\frac{a}{2} + \frac{π}{4}) da) {IBP}$ $set \frac{a}{2} + \frac{π}{4} = y \Rightarrow da = 2 dy$ $I = \frac{1}{4} (alntan (y) - 2 \int lntan (y) dy)$ $Ω = \int lntan (y) dy {\tan (y) = \frac{1 - e^{- 2 iy}}{1 + e^{- 2 iy}}}$ $Ω = \int \ln (1 - e^{- 2 iy}) dy - \int \ln (1 + e^{- 2 iy}) dy$ $set w = e^{- 2 iy} \Rightarrow dw = \frac{2 w}{i} dy$ $Ω = \frac{i}{2} \int \frac{\ln (1 - w)}{w} dw - \frac{i}{2} \int \frac{\ln (1 + w)}{w} dw$ $Ω = - \frac{i}{2} {Li}_{2} (w) - \frac{i}{2} (- {Li}_{2} (- w)) + C$ $Ω = \frac{i}{2} ({Li}_{2} (- e^{- 2 iy}) - {Li}_{2} (e^{- 2 iy}))$ $I = \frac{1}{4} (alntan (y) - i ({Li}_{2} (- e^{- 2 iy}) - {Li}_{2} (e^{- 2 iy})))$ $I = \frac{1}{2} xlntan (x + \frac{π}{4}) - \frac{i}{4} ({Li}_{2} (- e^{- i (2 x + \frac{π}{2})}) - {Li}_{2} (e^{- i (2 x + \frac{π}{2})})) + C$ $\int_{0}^{1} xsec (2 x) dx = (\frac{1}{2} lntan (\frac{4 + π}{4}) - \frac{i}{4} {Li}_{2} (- e^{- i (\frac{4 + π}{2})}) + \frac{i}{4} {Li}_{2} (e^{- i (\frac{4 + π}{2})})) + \frac{i}{4} ({Li}_{2} (- e^{- \frac{i π}{2}}) - {Li}_{2} (e^{- \frac{i π}{2}}))$ $\int_{0}^{1} xsec (2 x) dx = \infty$ $The integral diverges .$
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