# 02_20101427_13 - TAHSINUL HAQUE DHRUBO.pdf - MAT120...

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MAT120 Assignment SET-02 TAHSINUL HAQUE DHRUBO ID: 20101427 Sec:13 13th March 2021 1
1 Answer of the Question No-01 + Z 0 1 x ( x + 1) dx Let’s, x=y 2 dx=2ydy if x=0, y=0 and x=+ , y = + = + R 0 2 y ( y 2 +1) dy = + R 0 1 y 2 +1 dy = 2[ tan - 1 y ] 0 = 2( π 2 - 0) = π ( Ans. ) 2
2 Answer of the Question No-02 Z sec n x dx = R sec n - 2 xsec n x dx Let 0 s, u = sec n - 2 x or, du dx = ( n - 2) sec n - 3 secxtanx or, du = ( n - 2) sec n - 2 tanxdx Again, dv=sec 2 x v=tanx I n = uv - R vdu = sec n - 2 xtanx - R ( n - 2) sec n - 2 tanx tanx dx = secn - 2 xtanx - ( n - 2) R secn - 2 xtanx dx = sec n - 2 xtanx - ( n - 2) R sec n - 2 ( sec 2 x - 1) dx = sec n - 2 x tanx - ( n - 2) R sec n - 2 xsec 2 xsec n - 2 x dx = sec n - 2 xtanx - ( n - 2)( R sec n x - sec n - 2 x ) dx I n = sec n - 2 xtanx - ( n - 2)[ I n - I n - 2 ] I n + ( n - 2) In = sec n - 2 x tanx - ( n - 2) R sec n - 2 x dx I n = sec n - 2 x tanx n - 1 + n - 2 n - 1 R sec n - 2 xdx Now, R sec 5 x dx = sec 3 tanx 4 + 3 4 R sec 3 x dx = sec 3 tanx 4 + 3 4 [ secx tanx 2 + 2 3 R secx dx ] = sec 3 tanx 4 + 3 4 [ secx tanx 2 + 2 3 [ R secx ( secx + tanx ) dx secx + tanx ]] = sec 3 tanx 4 + 3 4 [ secx tanx 2 + 2 3 [ ln | secx + tanx | ]] + c ( Ans. ) 3
3 Answer of the Question No-03 Z 2 x 2 - 10 x + 4 ( x + 1)( x + 3) 2 dx 2 x 2 - 10 x +4 ( x +1)( x +3) 2 = A x +1 + B x - 3 + C ( x - 3) 2 2 x 2 - 10 x + 4 = A ( x - 3) 2 + B ( x - 3)( x + 1) + C ( x + 1) Let, s x = 3 , A (0) + B (0) + C (4) = - 8 C = - 2 Let 0 s, x = - 1 , 16 A + B (0) + C (0) = 16 A = 1 Let 0 s.x = 0 , 9 A - 3 B + C = 4 B = 1 After getting A, BC, R 1 x +1 + 1 x - 3 - 2 ( x - 3) 2 dx = R 1 x +1 dx + R 1 x - 3 dx - R 2 ( x - 3) 2 dx = ln | x + 1 | + ln | x - 3 | + 2 x - 3 ( Ans. ) 4
4 Answer of the Question No-04 2 Z 2 1 x 2 p ( x 2 - 1) dx Let 0 s, x =