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Question

Evaluate: 13+2sinx+cosxdx

A
I=tan1[tan(x2)1]+c
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B
I=tan1[tan(x2)+1]+c
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C
I=tan1[tan(x2)2]+c
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D
None of these
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Solution

The correct option is A I=tan1[tan(x2)+1]+c
Let I=13+2sinx+cosxdx
put tanx2=t
x=2tan1t
dx=2dt(1+t2) and sinx=2t(1+t2)
cosx=(1t2)1+t2)
I=13+2×2t1+t2+1+t21+t2×21+t2dt
=2dt3+3t2+4t+1t2
=2dt(t+1)21+2
=dtt2+2t+2
=dt(t+1)21+2
=dt(t+1)2+12
=tan1(t+1)+c
Therefore,
I=tan1[tan(x2)+1]+c

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