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Question

Evaluate : secxtanxsec2x+1dx.

A
12tt2+112log[t+t2+1]. where t=secx
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B
12tt2+1+12log[t+t2+1]. where t=secx
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C
tt2+1+log[t+t2+1]. where t=secx
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D
12tt2+1+12log[tt2+1]. where t=secx
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Solution

The correct option is B 12tt2+1+12log[t+t2+1]. where t=secx
Let I=secxtanxsec2x+1dx

Put secx=tsecxtanxdx=dt

Therefore

I=t2+1dt=12tt2+1+12log[t+t2+1]

Hence, option 'B' is correct.

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