The integral ∫sec2x(secx+tanx)9/2 dx\displaystyle\int\frac{\sec^2x}{(\sec x+\tan x)^{9/2}}\,dx∫(secx+tanx)9/2sec2xdx equals (for some arbitrary constant KKK)
Official answer: C
Substitute t=secx+tanxt=\sec x+\tan xt=secx+tanx. Then dt=secx (secx+tanx) dx=tsecx dxdt=\sec x\,(\sec x+\tan x)\,dx=t\sec x\,dxdt=secx(secx+tanx)dx=tsecxdx, and since secx−tanx=1t\sec x-\tan x=\frac1tsecx−tanx=t1, secx=12(t+1t)\sec x=\frac12\left(t+\frac1t\right)secx=21(t+t1).
Rewrite.sec2x dx=secx⋅dtt=12(1+1t2)dt.\sec^2x\,dx=\sec x\cdot\frac{dt}t=\frac12\left(1+\frac1{t^2}\right)dt.sec2xdx=secx⋅tdt=21(1+t21)dt.
Integrate.12∫(t−9/2+t−13/2)dt=−17t−7/2−111t−11/2+K=−1t11/2{111+17t2}+K.\frac12\int\left(t^{-9/2}+t^{-13/2}\right)dt=-\frac17t^{-7/2}-\frac1{11}t^{-11/2}+K=-\frac1{t^{11/2}}\left\{\frac1{11}+\frac17t^2\right\}+K.21∫(t−9/2+t−13/2)dt=−71t−7/2−111t−11/2+K=−t11/21{111+71t2}+K.
(C)