SET-02_20101528_10 - Nazmul Hasan Oyon.pdf - Assignment1...

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Assignment1 Nazmul Hasan Oyon ID:20101528 Section:10 14th March 2021 1. R 0 1 x ( x +1) d x Evaluating the integral over the finite integral 0 < = x < = t gives, R t 0 1 x ( x +1) d x = 2 R t 0 1 1+ u 2 d u [ x = u, u 2 = x, 2 udu = dx ] = 2[ tan - 1 u ] t 0 = 2[ tan - 1 ( x )] t 0 = π 2. I = R sec n ( x )d x = R sec n - 2 xsec 2 x d x = sec n - 2 xtanx - ( n - 2) R sec n - 3 xsecxtan 2 x d x = sec n - 2 xtanx - ( n - 2) R sec n - 2 xtan 2 x d x = sec n - 2 xtanx - ( n - 2) R sec n - 2 x ( sec 2 x - 1)d x = sec n - 2 xtanx - ( n - 2) R sec n x d x + ( n - 2) R sec n - 2 x d x = sec n - 2 xtanx - ( n - 2) I n + ( n - 2) I n - 2 I n (1 + n - 2) = sec n - 2 xtanx + ( n - 2) I n - 2 I n ( n - 1) = sec n - 2 xtanx + ( n - 2) I n - 2 I n = sec n - 2 xtanx n - 1 + n - 2 n - 1 I n - 2 if n = 5 , then I 5 = R sec 5 x d x I 5 = 1 4 sec 3 xtanx + 3 4 R sec 3 x d x Again, R sec 3 x d x = secxtanx 2 + 1 2 R secx d x = secxtanx 2 + 1 2 R secx secx + tanx secx + tanx d x = secxtanx 2 + 1 2 R sec 2 x + secxtanx secx + tanx d x let u = secx + tanx du = ( sec 2 x + secxtanx ) dx = secxtanx 2 + 1 2 R 1 u d u + c = secxtanx 2 + 1 2 ln | u | + c 1
= secxtanx 2 + 1 2 ln | secx + tanx | + c I 5 = 1 4 sec 3 xtanx + 3 4 R sec 3 x d x = 1 4 sec 3 xtanx + 3 8 [ secxtanx + ln | secx + tanx | ] + c 3 .

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