(a) Given that f is a function such that the integrals exist,
use the substitution \(u = a - x\) to show that
$$\int _ { 0 } ^ { a } \mathrm { f } ( x ) \mathrm { d } x = \int _ { 0 } ^ { a } \mathrm { f } ( a - x ) \mathrm { d } x$$
Hence use symmetry of \(\mathrm { f } ( \sin x )\) on the interval \([ 0 , \pi ]\) to show that
$$\int _ { 0 } ^ { \pi } x \mathrm { f } ( \sin x ) \mathrm { d } x = \pi \int _ { 0 } ^ { \frac { \pi } { 2 } } \mathrm { f } ( \sin x ) \mathrm { d } x$$
(b) Use the result of (a)(i) to show that
$$\int _ { 0 } ^ { \frac { \pi } { 2 } } \frac { \sin ^ { n } x } { \sin ^ { n } x + \cos ^ { n } x } \mathrm {~d} x$$
is independent of \(n\), and find the value of this integral.
(c) (i) Prove that
$$\frac { \cos x } { 1 + \cos x } \equiv 1 - \frac { 1 } { 2 } \sec ^ { 2 } \left( \frac { x } { 2 } \right)$$
Hence use the results from (a) to find
$$\int _ { 0 } ^ { \pi } \frac { x \sin x } { 1 + \sin x } \mathrm {~d} x$$
(d) Find
$$\int _ { 0 } ^ { \pi } \frac { x \sin ^ { 4 } x } { \sin ^ { 4 } x + \cos ^ { 4 } x } \mathrm {~d} x$$