Integrate challenge 1

Calculus Level 5

0 1 sin x ( cos 2 x cos 2 π 5 ) ( cos 2 x cos 2 2 π 5 ) sin 5 x . d x \large{\displaystyle \int^{1}_{0} \frac{\sin x(\cos^2 x-\cos^2 \frac{\pi}{5} )(\cos^2 x-\cos^2 \frac{2\pi}{5} )}{\sin 5x}.dx}

The value of above integral is a b \large{\frac{a}{b}} . Find a + b a+b .


The answer is 17.

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1 solution

Sudeep Salgia
Jul 26, 2015

This question revolves around a couple of identities.

_Identity 1: _ r = 0 n 1 sin ( x + r π n ) = sin ( n x ) 2 n 1 \prod_{r=0}^{n-1} \sin \left( x + \frac{r \pi }{n} \right) = \frac{\sin (nx)}{2^{n-1}}

Proof:

Consider the equation, z n = e 2 n i x z^n = e^{2nix} . We can factorize the above equation using the concept of roots of unity.
z n e 2 n i x = r = 0 n 1 ( z e i ( 2 x + 2 π r n ) ) 1 e 2 n i x = r = 0 n 1 ( 1 e i ( 2 x + 2 π r n ) ) 2 sin ( n x ) sin ( n x ) i cos ( n x ) = r = 0 n 1 2 sin ( x + π n ) sin ( x + π n ) i cos ( x + π n ) 2 sin ( n x ) = 2 n r = 0 n 1 sin ( x + r π n ) \displaystyle \begin{array}{c}\\ z^n - e^{2nix} && = \prod_{r=0}^{n-1} \left( z - e^{i \left(2x + \frac{2 \pi r}{n}\right) } \right) \\ | 1 - e^{2nix} | && = \prod_{r=0}^{n-1} \bigg| \left( 1 - e^{i \left(2x + \frac{2 \pi r}{n}\right) } \right) \bigg| \\ 2 \sin (nx) \big| \sin (nx) -i \cos (nx) \big| && = \prod_{r=0}^{n-1} 2 \sin \left(x + \frac{\pi}{n} \right) \bigg| \sin \left(x + \frac{\pi}{n} \right) -i \cos \left(x + \frac{\pi}{n}\right) \bigg| \\ 2 \sin (nx) && = 2^{n} \prod_{r=0}^{n-1} \sin \left( x + \frac{r \pi }{n} \right) \\ \end{array}

And the identity follows.

Identity 2: cos 2 A cos 2 B = sin ( B + A ) sin ( B A ) \cos ^2 A - \cos ^2 B = \sin (B+A) \sin (B - A)

Proof:

sin ( B + A ) sin ( B A ) = ( sin A cos B + cos A sin B ) ( cos A sin B sin A cos B ) = cos 2 A sin 2 B sin 2 A cos 2 B = cos 2 A cos 2 A cos 2 B sin 2 A cos 2 B = cos 2 A cos 2 B \displaystyle \begin{array}{c}\\ \sin (B+A) \sin (B - A) && = ( \sin A \cos B + \cos A \sin B)( \cos A \sin B - \sin A \cos B) \\ && = \cos^2 A \sin^2 B - \sin^2 A \cos^2 B \\ && = \cos^2 A - \cos^2 A \cos^2 B - \sin^2 A \cos^2 B \\ && = \cos ^2 A - \cos ^2 B \\ \end{array}


Using the second identity, we can write, sin x ( cos 2 x cos 2 y ) sin x ( cos 2 x cos 2 2 y ) \displaystyle \sin x ( \cos ^2 x - \cos ^2 y )\sin x ( \cos ^2 x - \cos ^2 2y )

= sin x sin ( x + y ) sin ( y x ) sin ( x + 2 y ) sin ( 2 y x ) = sin ( x 2 π 5 ) sin ( x π 5 ) sin ( x ) sin ( x + π 5 ) sin ( x + 2 π 5 ) = sin ( x ) sin ( x + π 5 ) sin ( x + 2 π 5 ) sin ( x + 3 π 5 ) sin ( x + 4 π 5 ) = sin ( 5 x ) 2 4 \displaystyle \begin{array}{c}\\ && = \sin x \sin (x + y ) \sin (y - x) \sin (x + 2y) \sin (2y - x) \\ && = \sin \left( x - \frac{2 \pi }{5} \right) \sin \left( x - \frac{ \pi }{5} \right) \sin (x) \sin \left( x + \frac{ \pi }{5} \right) \sin \left( x + \frac{2 \pi }{5} \right) \\ && = \sin (x) \sin \left( x + \frac{ \pi }{5} \right) \sin \left( x + \frac{2 \pi }{5} \right) \sin \left( x + \frac{3 \pi }{5} \right) \sin \left( x + \frac{ 4\pi }{5} \right) \\ && = \frac{ \sin(5x) }{2^4} \\ \end{array}

The last step follows from the first identity. Therefore,

0 1 sin x ( cos 2 x cos 2 y ) sin x ( cos 2 x cos 2 2 y ) sin ( 5 x ) d x \displaystyle \int_0^1 \frac{ \sin x ( \cos ^2 x - \cos ^2 y )\sin x ( \cos ^2 x - \cos ^2 2y )}{ \sin (5x) } \text{ d}x
= 0 1 sin ( 5 x ) 16 . 1 sin ( 5 x ) d x = 0 1 1 16 d x = 1 16 \displaystyle = \int_0^1 \frac{ \sin (5x)}{16}. \frac{1}{ \sin (5x) } \text{ d}x = \int_0^1 \frac{1}{16} \text{ d}x = \frac{1}{16}

Compare the answer to find the values of a a and b b and add them up to get the final answer as 17 \displaystyle \boxed{17} .

Moderator note:

What is the quick way to generalize the trigonometric identity that is stated?

IE Prove that for n = 2 k + 1 n = 2k+1 , sin n θ = sin θ × i = 1 k ( cos i π 2 k + 1 cos θ ) × i = 1 k ( cos i π 2 k + 1 + cos θ ) \sin n \theta = \sin \theta \times \prod_{i=1}^k ( \cos \frac{i \pi}{2k+1 } - \cos \theta) \times \prod_{i=1}^k ( \cos \frac{i \pi}{2k+1 } + \cos \theta) .

Nice Question ! @Tanishq Varshney

Sudeep Salgia - 5 years, 10 months ago

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