02_19201110_10 - MAHJABIN CHOWDHURY.pdf - Weekly Assignment...

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Weekly Assignment: 01 Name : Mahjabin Chowdhury ID : 19201110 Section : 10 Set: 02
Ans.to.the.Q.No:01 R + 0 dx x ( x +1) Let, x = u x = u 2 du = 1 2 x dx du = 1 2 u dx 2 udu = dx R + 0 dx x ( x +1) = lim k →∞ R k 0 dx x ( x +1) = lim k →∞ R k 0 2 u u ( u 2 +1) du = lim k →∞ R k 0 2 u 2 +1 du = lim k →∞ [2tan - 1 u ] k 0 = lim k →∞ [2tan - 1 x ] k 0 = lim k →∞ (2tan - 1 k - 2tan - 1 0) = 2 * π 2 = π So, it is convergent. (Answer.)
Ans.to.the.Q.No:02 R sec n x dx Let, I n = R sec n x dx ............................... (01) = R sec n - 2 x sec 2 x dx = sec n - 2 x R sec 2 x dx - R [ d dx ( sec n - 2 x ) R sec 2 x dx ] dx = sec n - 2 x tanx - R ( n - 2) sec n - 3 x ( secx tanx ) tanx dx = sec n - 2 x tanx - ( n - 2) R sec n - 2 x tan 2 x dx = sec n - 2 x tanx - ( n - 2) R sec n - 2 x ( sec 2 x - 1) dx = sec n - 2 x tanx - ( n - 2) R ( sec n x - sec n - 2 x ) dx = sec n - 2 x tanx - ( n - 2) R sec n x dx + ( n - 2) R sec n - 2 x dx = sec n - 2 x tanx - ( n - 2) I n + ( n - 2) I n - 2 ......................... [ from equation 01] I n = sec n - 2 x tanx - ( n - 2) I n + ( n - 2) I n - 2 I n (1 + n - 2) = sec n - 2 x tanx + ( n - 2) I n - 2 I n = sec n - 2 x tanx n - 1 + ( n - 2) ( n - 1) I n - 2 I n = sec n - 2 x tanx n - 1 + ( n - 2) ( n - 1) I n - 2 + C intsec n x dx = sec n - 2 x tanx n - 1 + ( n - 2) ( n - 1) I n - 2 + C Given, R sec 5 x dx By using reduction formula,
R sec 5 x dx = sec 5 - 2 x tanx 5 -

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