Page 29 - 12 Math 3
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31.. IQntueagdraratitoicn Equations eeLLeeaarrnn..PPuunnjjaabb
= ∫sin2 x dx - ∫sin2 x cos2 x dx
∫1 - cos 2x dx - ∫sin2 x cos2 x dx (I)
2
Integrating ∫ sin2 x cos2 x dx by parts, we have
∫ ∫sin2 x cos2 x dx = cos x sin2 x cos x dx
= cos x  sin3 x - sin3 x × (- sin x) dx [ï‘ If f (x) = cos x and
 3  g'(x) = sin2 x cos x.
ï£ ï£¸ ∫3
∫=1 cos x sin3 x + 1 sin4 x dx ..... (II) then f '(x) = - sin x
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and g(x) = sin2 sin3 x 

3 
Putting the value of ∫ sin2 x cos2 x dx in (I), we obtain,
∫sin4 x dx =∫ï£ï£« 12 - cos 2x  dx -  1 cos x sin3 x + 1 ∫ sin 4 x dx
2   3 3
=12 ∫1 dx - 1 ∫ cos 2x dx - 1 cos x sin3 x - 1 ∫ sin 4 x dx
2 3 3
∫or ï£ï£¬ï£«1 +1 sin4 x dx= 1 × - 1  sin 2x  + c1 - 1 cos x sin3 x
3  2 2 ï£¬ï£ 2  3
∫ sin4 x=dx 3 1 × -1 sin 2x - 1 cos x sin3 x + c 
4  2 4 3 
=3 x - 3 sin 2x - 1 cos x sin3 x + c where c =34 c1
8 16 4
Example 10. Evaluate ∫ e x (1 + sin x ) dx.
1 + cos x
Solution: e x (1 + sin x ) ∫ ex ï£ï£¬1 + 2 sin x cos x  ï£¯ï£°ï£®ï‘ 1 + cosx = 1 + 2cos2 x - 1
cos x 2 2  2
∫ 1+ dx dx
2cos2 x
2
version: 1.1
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