Question

${\int }_0^{\pi /4}{\left(tan\right)}^6(x-[x])+{\left(tan\right)}^4(x-[x])dx$ is equal to :-
(where $[.]$ is $G.I.F.$ )

Answer 1

Answer: 0.20

$\because 0<x\le \frac{\pi }{4}\Rightarrow [x]=0$
$\because I={\int }_0^{\pi /4}\left({\left(tan\right)}^{6}x+{\left(tan\right)}^{4}x\right)dx$
$={\int }_0^{\pi /4}ta{n}^{4}x\cdot se{c}^{2}xdx=\frac{1}{5}=0.20$