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$If P = \int \frac{\sec^{2} x}{(\sec x + \tan x)^{10}} d x then P is equal to$

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a

118secx+tanx9−122secx+tanx11+K

b

118secx+tanx9+111secx+tanx11+K

c

−118secx+tanx9−122secx+tanx11+K

d

−118secx+tanx9+122secx+tanx11+K

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detailed solution

Correct option is C

Let sec⁡x+tan⁡x=t⇒sec⁡xtan⁡x+sec2⁡xdx=dt⇒sec⁡xdx=dtt And sec⁡x+tan⁡x=t→(1)sec⁡x−tan⁡x=1t→(2)(1)+(2)2sec⁡x=t+1tsec⁡x=121+t2t∴P=∫12t2+1tdttt10=12∫t2+1t12dt=12∫1t10+1t12dt=12−19t9−111t11+K=−118[sec⁡x+tan⁡x]9−122[sec⁡x+tan⁡x]11+K

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