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Question

The integral sec2x(secx+tanx)9/2dx equals

A
(secxtanx)11/2{111+17(secxtanx)2}+K
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B
1(secx+tanx)11/2{111+17(secx+tanx)2}+K
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C
1(secx+tanx)11/2{111+17(secx+tanx)2}+K
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D
(secxtanx)11/2{111+17(secxtanx)2}+K
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Solution

The correct options are
A (secxtanx)11/2{111+17(secxtanx)2}+K
C 1(secx+tanx)11/2{111+17(secx+tanx)2}+K
Let secx+tanx=t
secx(secx+tanx)dx=dt
Also, 1t=secxtanx (sec2xtan2x=1)
2secx=t+1t
secx=12(t+1t) (i)
Then, sec2x(secx+tanx)9/2dx
=sec2xt9/2.dttsecx
=secxt11/2dt
=12t+1tt11/2dt [From (i)]
=12dtt9/2+12dtt13/2
=17t7/2111t11/2+K
=1(secx+tanx)11/2{111+17(secx+tanx)2}+K
=(secxtanx)11/2{111+17(secxtanx)2}+K [secx+tanx=1secxtanx]

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