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Q.

∫dta2sin2⁡t+b2cos2⁡t is equal to

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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∫sec2⁡tdttan2⁡t+ba2 Put tan⁡t=rsec2tdt=dr Use ∫1a2+x2dx=1atan−1⁡xa+c=1a2∫drr2+ba2=1a21batan−1⁡rba+k=1abtan−1⁡atan⁡tb+k
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