prove that Σ_(k=0) ^(k=n) cos^2 (kx)= ((n+1)/2) + ((sin((n+1)x)cos(nx))/(2 sinx)) x from R−{ kπ.kεZ}then find the value of integral ∫


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Question Number 26242 by abdo imad last updated on 22/Dec/17
$prove that Σ_(k=0) ^(k=n) cos^2 (kx)= ((n+1)/2) + ((sin((n+1)x)cos(nx))/(2 sinx)) x from R−{ kπ.kεZ}then find the value of integral ∫_0 ^π ((sin((n+1)x)cos(nx))/(sinx))dx$
$p r o v e t h a t \sum_{k = 0}^{k = n} {c o s}^{2} (k x) = \frac{n + 1}{2} + \frac{s i n ((n + 1) x) c o s (n x)}{2 s i n x}$ $x f r o m R - {k π . k ε Z} t h e n f i n d t h e v a l u e o f i n t e g r a l$ $\int_{0}^{π} \frac{s i n ((n + 1) x) c o s (n x)}{s i n x} d x$
Commented by abdo imad last updated on 28/Dec/17

$\sum_{k = 0}^{k = n} {c o s}^{2} (k x) = \frac{1}{2} \sum_{k = 0}^{k = n} (1 + c o s (2 k x))$ $= \frac{n + 1}{2} + \frac{1}{2} \sum_{k = 0}^{n} c o s (2 k x) b u t \sum_{k = 0}^{k = n} c o s (2 k x) = R e (\sum_{k = 0}^{k = n} e^{i 2 k x})$ $= R e (\frac{1 - e^{i 2 (n + 1) x}}{1 - e^{i 2 x}}) b u t \frac{1 - e^{i 2 (n + 1) x}}{1 - e^{i 2 x}} = \frac{1 - c o s 2 (n + 1) x - i s i n (2 (n + 1) x)}{1 - c o s (2 x) - i s i n (2 x)}$ $= \frac{2 {s i n}^{2} (n + 1) x - 2 i s i n (n + 1) x c o s (n + 1) x}{2 {s i n}^{2} x - 2 i s i n x c o s x}$ $= \frac{- i s i n (n + 1) x (c o s (n + 1) x + i s i n (n + 1) x)}{- i s i n x (c o s x + i s i n x)}$ $= \frac{s i n (n + 1) x}{s i n x} e^{i (n + 1) x} e^{- i x} = \frac{s i n (n + 1) x}{s i n x} e^{i n x}$ $= \frac{s i n (n + 1) x}{s i n x} c o s (n x) + i \frac{s i n (n + 1) x s i n (n x)}{s i n x}$ $\Rightarrow R e (\sum_{k = 0}^{k = n} e^{i 2 k x}) = \frac{s i n (n + 1) x c o s (n x)}{s i n x}$ $\Rightarrow \sum_{k = 0}^{k = n} {c o s}^{2} (k x) = \frac{n + 1}{2} + \frac{s i n (n + 1) x c o s (n x)}{2 s i n x}$ $\int_{0}^{π} \frac{s i n (n + 1) x c o s (n x)}{2 s i n x} d x = \sum_{k =>}^{n} \int_{0}^{π} {c o s}^{2} (k x) d x - \frac{(n + 1) π}{2}$ $= \frac{1}{2} \sum_{k = 0}^{k = n} \int_{0}^{π} (1 + c o s (2 k x)) d x - \frac{(n + 1) π}{2}$ $= \frac{1}{2} \sum_{k = 0}^{k = n} \int_{0}^{π} c o s (2 k x) d x = \frac{π}{2} + \sum_{k = 1}^{k = n} {[\frac{1}{2 k} s i n (2 k x)]}_{k = 0}^{k = π =}$ $= \frac{π}{2} + 0 = \frac{π}{2} \Rightarrow \int_{0}^{π} \frac{s i n (n + 1) x c o s (n x)}{s i n x} d x = π$
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