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dxa2sin2x+b2cos2x is equal to 

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a
1abtan−1⁡atan⁡xb+C
b
abtan−1⁡atan⁡xb+C
c
abtan−1⁡atan⁡xb-C
d
None of the above

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detailed solution

Correct option is A

∫dxa2sin2⁡x+b2cos2⁡x=∫sec2⁡xdxa2tan2⁡x+b2=1a2∫dtt2+ba2 where, t=tan⁡x=1a2⋅abtan−1⁡tb/a+C=1abtan−1⁡a(tan⁡x)b+C


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