One Two Three Four Five Present!

$The\quad integral\quad can\quad be\quad written\quad as\quad \\ \\ \int _{ \frac { \pi }{ 4 } }^{ \frac { \pi }{ 2 } }{ \frac { 2\cos { \frac { 9x }{ 2 } .\cos { \frac { x }{ 2 } } } .dx }{ 1-2\left( 2\cos ^{ 2 }{ \frac { 3x }{ 2 } } -1 \right) } } \\ \\ Multiplying\quad the\quad numerator\quad and\quad denominator\quad by\quad \cos { \frac { 3x }{ 2 } } \\ \\ \int _{ \frac { \pi }{ 4 } }^{ \frac { \pi }{ 2 } }{ \frac { 2\cos { \frac { 9x }{ 2 } } .\cos { \frac { x }{ 2 } } .\cos { \frac { 3x }{ 2 } } }{ 3\cos { \frac { 3x }{ 2 } } -4\cos ^{ 3 }{ \frac { 3x }{ 2 } } } .dx } \\ \\ Now\quad using\quad \cos { 3\theta } =4\cos ^{ 3 }{ \theta } -3\cos { \theta } \quad we\quad get\\ \\ \int _{ \frac { \pi }{ 4 } }^{ \frac { \pi }{ 2 } }{ \frac { 2\cos { \frac { 9x }{ 2 } } .\cos { \frac { x }{ 2 } } .\cos { \frac { 3x }{ 2 } } }{ -\cos { \frac { 9x }{ 2 } } } .dx } \\ \\ -\int _{ \frac { \pi }{ 4 } }^{ \frac { \pi }{ 2 } }{ 2\cos { \frac { x }{ 2 } } } .\cos { \frac { 3x }{ 2 } .dx } \\ -\int _{ \frac { \pi }{ 4 } }^{ \frac { \pi }{ 2 } }{ (\cos { x } +\cos { 2x } } ).dx\\ \\ Now\quad the\quad integral\quad can\quad be\quad evaluated.\\$ $Which\quad comes\quad out\quad to\quad be\quad \frac { 1 }{ \sqrt { 2 } } -\frac { 1 }{ 2 } \\ Hence\quad the\quad answer\quad is\quad 2+\sqrt { 2 } .$

Using the identity $cosA+cosB=2cos\frac{A+B}{2}\ cos\frac{A-B}{2}\$ , $cos5x+cos4x$ can be expressed as $2cos\frac { 9 }{ 2 } \ x cos\frac{1}{2}\ x$ . With this identity $\frac{cos3x}{cox}=2cos2x-1$ , the denominator can be expressed as $-\frac{cos\frac{9}{2}x}{cos\frac{3}{2}x}$ , where the angle is replaced by $\frac{3x}{2}$ .

Bringing them together, we get $\int _{ \pi /4 }^{ \pi /2 }{ \frac{2cos\frac { 9 }{ 2 } \ x cos\frac{1}{2}\ x}{ -\frac{cos\frac{9}{2}x}{cos\frac{3}{2}x}}dx }$ .

After simplification, we obtain $\int _{ \pi /4 }^{ \pi /2 }{ -2cos\frac { 3x }{ 2 } cos\frac { x }{ 2 }dx }$ which can be rewritten as $-\int _{ \pi /4 }^{ \pi /2 }({ cos2x } +cosx )dx$ $=-{ \left[ \frac { sin 2x }{ 2 } +sin x \right] }_{ \pi /4 }^{ \pi /2 }\\ =\quad -1+\frac { 1 }{ 2 } +\frac { 1 }{ \sqrt { 2 } } \\ =\frac { 1 }{ \sqrt { 2 } } -\quad \frac { 1 }{ 2 }$

Therefore, $a+b=3.414$