Page 7 - CH-7-SLRC-HOME ASSIGNMENT

dx
6. Evaluate∫
2
x −6x+13
2
2
Soln: x − 6x + 13 = (x − 3) + 4
dx dx dx
So ∫ = ∫ = ∫
2
2
2
x −6x+13 (x−3) +4 (x−3) +2 2
Put x − 3 = t, then dx = dt
dx dx 1 −1 t 1 −1 x−3
Therefore ∫ = ∫ = tan + C = tan + C
2
2
x −6x+13 t +2 2 2 2 2 2
dx
7. Evaluate ∫
(x+1)(x+2)
1 A B
Soln: = + where A and B are to be determined.
(x+1)(x+2) x+1 x+2
This gives 1 = A(x + 2) + B(x + 1)
Equating the coefficients of x and the constant term we get A+B=0 and 2A+B=1
Solving these equations we get A = 1 and B = -1
1 1 −1
Thus the integrand is given by = +
(x+1)(x+2) x+1 x+2
dx dx dx x+1
Therefore ∫ = ∫ − ∫ = log|x + 1| − log|x + 2| + C = log| | + C
(x+1)(x+2) x+1 x+2 x+2
8. Evaluate ∫ logxdx
d
∫ logxdx = ∫ logx. 1dx = logx ∫ dx − ∫ (logx)∫ dx
dx
1
= (logx)x − ∫ x dx = xlogx − x + C
x
2
9. Evaluate ∫ √x + 2x + 5dx
2
2
∫ √x + 2x + 5 dx = ∫ √(x + 1) + 4dx
Put x + 1 = y so that dx = dy
2
2
2
Then ∫ √x + 2x + 5 dx = ∫ √y + 2 dy
1 4
2
2
= y√y + 4 + log|y + √y + 4| + C
2 2
1
2
2
= (x + 1)√x + 2x + 5 + 2log|x + 1 + √x + 2x + 5| + C
2
π
3
4
10. Evaluate ∫ sin 2tcos2tdt
0
1
Put sin2t = u so that 2cos2tdt = du or cos2tdt = du
2
1
1
1
4
3
4
3
So ∫ sin 2tcos2tdt = ∫ u du = u = sin 2t = F(t)
2 8 8
π
4 1 π 1
4
3
∫ sin 2tcos2tdt = [sin 4 − sin 0] =
0 8 2 8