$$In=$$∫tann⁡xdx then $$In+In+2$$ is equal to

Correct option is 2tann+1xn+1+C

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$I_{n} = \int \tan^{n} x d x then I_{n} + I_{n + 2} is equal to$

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tann−1xn−1+C

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detailed solution

Correct option is B

In+In+2=∫tannxdx+∫tann+2xdx=∫tann⁡xtan2⁡x+1dx=∫tann⁡xsec2⁡xdx Put tan⁡x=tsec2⁡xdx=dt=∫tndt=tn+1n+1=tann+1n+1+C

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In=∫tann⁡xdx then In+In+2 is equal to

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$I_{n} = \int \tan^{n} x d x then I_{n} + I_{n + 2} is equal to$