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dxa2cos2x+b2sin2x(a,b>0) is equal to

 

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a
sin−1⁡((a/b)tan⁡x)+C
b
tan−1⁡((b/a)tan⁡x)+C
c
1abtan−1⁡batan⁡x+C
d
none of these

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detailed solution

Correct option is C

Putting t=tan⁡x we get∫dxa2cos2⁡x+b2sin2⁡x=∫sec2⁡xdxa2+b2tan2⁡x=∫dta2+b2t2=1b2∫dtt2+a2/b2=1b2⋅batan−1⁡bat+C=1abtan−1⁡batan⁡x+C.


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