complex

No, its much simpler.

Notice that \cos \frac{(2k-1) \pi}{10} ; 1 \le k \le 5 are the five distinct roots

to the equation \cos 5\theta = 0.

Now you perhaps know that \cos n \theta for any n can be expanded as a polynomial in \cos \theta of degree n and leading coefficient 2^n-1

[This sequence of polynomials are named Chebyshev Polynomials.

See Trigonometric definition in http://en.wikipedia.org/wiki/Chebyshev_polynomials.

Otherwise notice that \cos \n \theta + \cos (n-2) \theta = 2 \cos (n-1) \theta \cos \theta) and use strong induction to prove this statement]

So, \cos 5 \theta = P(\cos \theta) = 2^{n-1} \prod_{k=1}^5 \left(\cos \theta - \cos \frac{(2k-1) \pi}{10} \right)

Hence the given expression equals \frac{\cos 5 \theta}{16}

and the integral \frac{\sin 5 \theta}{80} + C

(Note how shamelessly I have changed the answer :D)

i got it sir
\cos{n\theta }=\cos^n\theta \left(1-\binom{n}{2}\tan^{2}\theta +\binom{n}{4}\tan^{4}\theta+............ \right)\\ \cos{n\theta }=\left(\cos^n\theta -\binom{n}{2}\sin^{2}\theta \cos^{n-2}\theta +\binom{n}{4}\sin^{4}\theta \cos^{n-4}\theta +............ \right)\\ put \ \sin^2\theta=(1-\cos^2\theta ) \\ \text{we will get coefficeint of } \cos^n \theta =\frac{1+\binom{n}{1}+\binom{n}{2}+.......\binom{n}{n}}{2}\\\left(\text{taking in consideration that}\binom{n}{r}=\binom{n}{n-r} \right)\\ =2^{n-1}
prophet sir is it right ?
and i hpe the other degrees are also having same coefficient
as it is obvious from symmetry and the signs will oscillate

That's what I asked you to prove using induction.

Consider the sequence \{a_n\} given by a_n = \cos n \theta

From the trigonometric relation in the previous post, this sequence satisfies the equationa_n + a_{n-2} = 2 a_{n-1} a_1

Our hypothesis is two-fold:

1) a_n can be expresssed as a polynomial of degree n in a₁ for all n (recall that a_1 = \cos \theta)

2) The coefficient of \cos ^{n} \theta i.e a_1^n is 2^{n-1}

Now, we use strong induction to prove this. So, after verifying the statement for n=1, 2 let's assume that the above are true for all n≤k. Now lets prove the statement for n=k+1.

We have using the equation a_{k+1} + a_{k-1} = 2a_ka_1

Write this as a_{k+1} = 2a_ka_1- a_{k-1}. Its easy to see that RHS is a polynomial of degree n in a₁ with leading coefficient 2_k

From the induction hypothesis, a_k is a polynomial of degree k with leading coefficient 2^k-1. Of course a_k-1 is of degree k-1 and does not have any further role.

So its immediately apparent that our hypothesis is proved for n=k+1 and hence is true for all natural numbers.