If \[\int {{{dx} \over {{{\cos }^3}x\sqrt {2\sin 2x} }}} = {\left( {\tan x} \right)^A} + C{\left( {\tan x} \right)^B} + k,\]
where k is a constant of integration, then A + B +C equals :
Solutions · 1
Official solution
\[\int {{{dx} \over {{{\cos }^3}x\sqrt {2\sin 2x} }}} \]
= \[\int {{{dx} \over {{{\cos }^3}x\sqrt {4\sin x\cos x} }}} \]
= \[\int {{{dx} \over {2{{\cos }^4}x\sqrt {\tan x} }}} \]
Let tan x = t2
\[ \Rightarrow \] \,\,\,\[ sec2xdx = 2t dt
as sec2x = 1 + tan2x = 1 + t4
= \]\int {{{{{\sec }^4}x\,dx} \over {2\sqrt {\tan x} }}} =∫2tanxsec2x(sec2xdx)=∫2t(1+t4)2tdt=∫(1+t4)dt= t +5t5+ k
=tanx+51tan^{{5 \over 2}}\[x + k
By comparing with the given equation, we get
A = \]{1 \over 2}, B =25, C =51∴A + B + C =516