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Q.

In=∫tann⁡xdx then In+In+2 is equal to

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a

tann−1xn−1+C

b

tann+1xn+1+C

c

tannxn+C

d

tan2n+22n+2+C

answer is B.

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Detailed Solution

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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