Simple ?

By Using Complex Numbers :

Let $y\quad =\quad { e }^{ ix }\quad$ .

$2\cos { (nx) } =\quad { y }^{ n }+\cfrac { 1 }{ { y }^{ n } } \\ 2\cos { x } =\quad y+\cfrac { 1 }{ y } \\ { (2\cos { x } ) }^{ 5 }\quad =\quad { (y+\cfrac { 1 }{ y } ) }^{ 5 }\quad \\ { (2\cos { x } ) }^{ 5 }\quad =\quad { y }^{ 5 }\quad +\quad 5{ y }^{ 3 }\quad +\quad 10y\quad +\quad \cfrac { 10 }{ y } +\cfrac { 5 }{ { y }^{ 3 } } +\cfrac { 1 }{ { y }^{ 5 } } \\ \\ { (2\cos { x } ) }^{ 5 }\quad =\quad ({ y }^{ 5 }\quad +\cfrac { 1 }{ { y }^{ 5 } } )\quad +\quad 5({ y }^{ 3 }+\cfrac { 1 }{ { y }^{ 3 } } )\quad +\quad 10(y+\cfrac { 1 }{ y } )\\ \\ \cos ^{ 5 }{ x } =\quad \cfrac { 1 }{ 16 } (\cos { 5x } \quad +\quad 5\cos { 3x } \quad +\quad 10\cos { x } )\\ \\ \int { \cos ^{ 5 }{ x\quad dx } } \quad =\quad \int { \cfrac { 1 }{ 16 } (\cos { 5x } \quad +\quad 5\cos { 3x } \quad +\quad 10\cos { x } )dx }$ .

Now Compute This Simple Integral :)

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Credit of Solution : @megh choksi

I used a technique other than what was suggested:

$I(5) = \int_0^{\frac{\pi}{4}} \cos^5 x\: dx$

Since $\cos^2 x = 1 - \sin^2 x$ , we get:

$= \int_0^{\frac{\pi}{4}} (1 - \sin^2 x)^2 \cos x\: dx$

Let $u = \sin x$ . Thus $du = \cos x\: dx$ :

$= \int_0^{\frac{\sqrt{2}}{2}} (1 - u^2)^2 \: du$ $= \int_0^{\frac{\sqrt{2}}{2}} (1 - 2u^2 + u^4)\: du$ $= \left.\left(u - \frac{2u^3}{3} + \frac{u^5}{5}\right)\right|_0^{\frac{\sqrt{2}}{2}}$ $= \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{6} + \frac{\sqrt{2}}{40} = \frac{43\sqrt{2}}{120}$

Thus, $a = 43,\: b = 2,$ and $c = 120$ , and $a + b + c = \boxed{165}$ .

As the title says yes it is simple.

Simply form the recursion and solve it.

$\displaystyle {I}_{n}=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ { cos }^{ n }(x)dx } =\int _{ 0 }^{ \frac { \pi }{ 4 } }{ { cos }^{ n-1 }(x).cos(x)dx }$

Applying intgration by parts we get :

$\displaystyle {I}_{n}={ cos }^{ n-1 }(x).sin(x)\overset { \frac { \pi }{ 4 } }{ \underset { 0 }{ | } } -(n-1)\int _{ 0 }^{ \frac { \pi }{ 4 } }{ { cos }^{ n-2 }(x).{ sin }^{ 2 }(x)dx }$

$\displaystyle {I}_{n}={ cos }^{ n-1 }(x).sin(x)\overset { \frac { \pi }{ 4 } }{ \underset { 0 }{ | } } -(n-1)\int _{ 0 }^{ \frac { \pi }{ 4 } }{ { cos }^{ n-2 }(x).(1-{cos}^{2}(x))dx }$

${ I }_{ n }={ (\frac { 1 }{ \sqrt { 2 } } ) }^{ n }+(n-1){ I }_{ n-2 }-(n-1){ I }_{ n }$

Simplifying this we get :

${ I }_{ n }={ \frac { 1 }{ n } (\frac { 1 }{ \sqrt { 2 } } ) }^{ n }+\frac { n-1 }{ n } { I }_{ n-2 }$

Also we have the initial values ${I}_{1}=\frac{1}{\sqrt{2}},{I}_{0}=\frac{\pi}{4}$

Solving the recursion we have ${I}_{5}=\frac{43\sqrt{2}}{120}$

Y not simply sub sinx=t