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Q.

Evaluate ∫tan⁡xa+btan2⁡xdx

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a

=12(b+a)log⁡acos2⁡x+bsin2⁡x+C

b

=12(b−a)log⁡acos2⁡-bsin2⁡x+C

c

=1(b−a)log⁡acos2⁡x+bsin2⁡x+C

d

=12(b−a)log⁡acos2⁡x+bsin2⁡x+C

answer is D.

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Detailed Solution

I=∫tan⁡xa+btan2⁡xdx=∫sinxcosxa+bsin2⁡xcos2⁡xdx=∫sin⁡xcos⁡xacos2⁡x+bsin2⁡xdx=12∫sin⁡2xacos2⁡x+bsin2⁡xdx Put acos2⁡x+bsin2⁡x=t∴ (b−a)sin⁡2xdx=dt∴ l=12(b−a)∫dtt      =12(b−a)loge⁡|t|+C      =12(b−a)log⁡acos2⁡x+bsin2⁡x+C
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