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a
1atan−1atantb+k
b
−1abtan−1batant+c
c
1abtan−1batant+k
d
1abtan−1abtant+k
answer is D.
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Detailed Solution
∫1a2sin2t+b2cos2tdt=∫1a2cos2tdttan2t+b2a2=1a2∫sec2tdttan2t+ba2 Put tant=rsec2tdt=dr Use ∫1a2+x2dx=1atan−1xa+c=1a2∫drr2+ba2=1a21batan−1rba+k=1abtan−1atantb+k