$$I=$$∫$$sec3$$⁡xdx is equal to x∈$$0$$,π$$2$$

Correct option is 2tanxsecx2+12logsecx+tanx+C

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$I = \int \sec^{3} x d x is equal to [x \in (0, \frac{π}{2})]$

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tanxsecx2+12logsecx−tanx+C

tanxsecx2+12logsecx+tanx+C

tanxsecx2−12logsecx+tanx+C

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detailed solution

Correct option is B

Let I=∫sec3⁡xdx=∫sec⁡xsec2⁡xdxI=∫1+tan2⁡xsec2⁡xdx Put tan⁡x=t⇒sec2⁡xdx=dt=∫1+t2dt=t21+t2+12log⁡t+t2+1+C=tan⁡x21+tan2⁡x+12log⁡tan⁡x+tan2⁡x+1+C=tan⁡xsec⁡x2+12log⁡|tan⁡x+sec⁡x|+C

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I=∫sec3⁡xdx is equal to x∈0,π2

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$I = \int \sec^{3} x d x is equal to [x \in (0, \frac{π}{2})]$