The Ideal Integral

Let us assume

$Ι = \int_0^\infty e^{-x}(sin^{25}x)dx$

By using the method of integration by-parts

$I= \left . -e^{-x}sin^{25}x\right|_0^\infty$ $+$ $25\int_0^\infty e^{-x}(sin^{24}x)(cos x)dx$

$I= 0 + 25[\left . (-e^{-x}sin^{24}x)(cos x)\right|_0^\infty + \int_0^\infty e^{-x}(24(sin^{23}x)(cos^{2}x) - sin^{25}x)]dx$

$I=25\int_0^\infty e^{-x}[24(sin^{23}x)(1-sin^{2}x) - sin^{25}x]dx$

$I=25\int_0^\infty e^{-x}[24sin^{23}x - 25sin^{25}x]dx$

$I=25\int_0^\infty e^{-x}24sin^{23}xdx - 625\int_0^\infty e^{-x}sin^{25}xdx$

$I=600\int_0^\infty e^{-x}sin^{23}xdx - 625I$

$626I=600\int_0^\infty e^{-x}sin^{23}xdx$

The given question can be written as

$=\frac{626I}{\int_0^\infty e^{-x}sin^{23}xdx}$

$=600\frac{\int_0^\infty e^{-x}sin^{23}xdx}{\int_0^\infty e^{-x}sin^{23}xdx}$

$=600$