Page 6 - CH-7-SLRC-HOME ASSIGNMENT

 

 
 

(iv) sina sin(a h) sin(a 2h) ... sin[a (n 1)h]
 1 


sin  nh 
  2   sin a 1 (n 1)h

 1  2
sin  h 
 2 
(v) cosa cos(a h) cos(a 2h) ... cos[a (n 1)h]       
 1 


sin  nh 
  2 1    cos a 1 (n 1)h

sin   h  2
 2 
EXAMPLES

1. Write an anti derivative of cos2x using the method of inspection.
d
Soln: we know that sin2x = 2cos2x
dx
1 d d 1
⇒ cos2x = sin2x = ( sin2x)
2 dx dx 2
1
So the anti derivative of cos2x is sin2x
2
1−sinx
2. Find the integral ∫ dx
2
cos x
1−sinx 1 sinx
Soln: ∫ dx = ∫ dx − ∫ dx
2
2
2
cos x cos x cos x
2
= ∫ sec xdx − ∫ tanxsecxdx
= tanx − secx + c
2
3. Evaluate ∫ 2xsin(x + 1)dx
2
Put x + 1 = t, so that 2xdx = dt
2
2
Therefore ∫ 2xsin(x + 1)dx = ∫ sintdt = −cost + C = −cos (x + 1) + C
4. Evaluate ∫ sin2xcos3xdx
1
Soln: ∫ sin2xcos3xdx = [∫ sin5xdx − ∫ sinxdx]
2
1 1 1 1
= [− cos5x + cosx] + C = − cos5x + cosx + C
2 5 10 2

dx
5. Evaluate ∫
√2x−x 2
dx dx
Soln: ∫ = ∫
√2x−x 2 √1−(x−1) 2
Put x − 1 = t. Then dx = dt
dx dt
−1
−1
Therefore ∫ = ∫ = sin t + C = sin (x − 1) + C
√2x−x 2 √1−t 2