The intergral ∫sec2x(secx+tanx)92dx equal (for some arbitrary constant k)
A
−1(secx+tanx)112{111−17(secx+tanx)2}+K
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B
1(secx+tanx)112{111−17(secx+tanx)2}+k
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C
−1(secx+tanx)112{111+17(secx+tanx)2}+k
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D
1(sexx+tanx)112{111+17(secx+tanx)2}+k
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Solution
The correct option is C−1(secx+tanx)112{111+17(secx+tanx)2}+k I=∫sec2x(secx+tanx)9/2dxLet secx+tanx=t⇒secx−tanx=1t⇒secx=12(t+1t)Also secx(secx+tanx)dx=dt⇒secxdx=dtt∴I=12∫(t+1t)dtt92.t=12∫(t−9/2+t−13/2)dt=12[t−9/2+1−92+1+t−13/2+1−132+1]+K=−17t−7/2−111t−112+k=−17t7/2−111t11/2+K=−1t11/2(111+t27)+K=−1(secx+tanx)11/2{111+17(secx+tanx)2}+K