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Question : 77 of 160
Marks:
+1,
-0
Solution:
∫dx+∫dy ∫dx+∫dy ∫(√1+x−√1−x)dx+∫y√y+1−y√y−1dy ++∫y√y+1−y√y−1dy . . . (i)
For
∫y√y+1dy⇒y+1=t2 On differentiate,
dy=2tdt ∫(t2−1)t(2tdt) =2∫(t4−t2)dt=2(−)+C For
∫y√y−1dy y−1=t2⇒dy=2tdt ∫(t2+1)t⋅2tdt =2∫t4+t2dt=t5+t3+C Now, from Eq. (i)
++− −− (y+1)3∕2⏟(−)f(y)+(y−1)3∕2⏟[−]g(y) A= and
B= (−)−(−)=
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