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I tried to expand $\sin(7x)$ into $\sin x,\cos x $ terms but integral is getting pretty lengthy and cumbersome , how do i approach in a simpler way ?

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  • $\begingroup$ This looks nasty, how did you run into this monster? $\endgroup$ – gt6989b Oct 12 '15 at 17:02
  • $\begingroup$ @gt6989b just a casual class test , but wolfram gives a simple 1/6 as the answer . $\endgroup$ – Sujith Sizon Oct 12 '15 at 17:08
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Observe that $$\sin 7x = \sin 6x \cos x + \cos 6x \sin x,$$ hence $$\cos^5 x \sin 7x = \sin 6x \cos^6 x + \sin x \cos^5 x \cos 6x.$$ Now here is the tricky part: if we let $u = \cos^6 x$, $v = \cos 6x$, we get $$u' = -6 \cos^5 x \sin x,$$ and $$v' = -6 \sin 6x.$$ So what does this suggest?

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  • $\begingroup$ Oh nice i also smelled some relation between power 5 and inner 6's but the substitution disnt cross my mind , thank you so much. $\endgroup$ – Sujith Sizon Oct 12 '15 at 17:17
  • $\begingroup$ Btw $u' = -6 \cos^5 x \sin x,$ $\endgroup$ – Sujith Sizon Oct 12 '15 at 17:26
  • $\begingroup$ @SujithZis Corrected. Now just divide by $-6$ to get the required relationship. $\endgroup$ – heropup Oct 12 '15 at 17:28
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No need to do any clever thinking :) Since \begin{align*} \cos^5(x) &= \left( \frac{e^{ix} + e^{-ix}}{2}\right)^5 \\ &= \frac{1}{32} \left[ \sum_{k=0}^{5} {5 \choose k} e^{kix}e^{-(5-k)ix}\right] \\ &= \frac{1}{32} \left[ \sum_{k=0}^{5} {5 \choose k} e^{(2k-5)ix}\right] \end{align*} and \begin{align*} \sin(7x) &= \frac{1}{2i} \left(e^{7ix} - e^{-7ix}\right) \end{align*} we have $$ \cos^5(x) \sin(7x) = \frac{1}{64i} \sum_{k=0}^5 {5 \choose k} \left[ e^{(2k+2)ix} - e^{(2k-12)ix} \right] $$ so \begin{align*} \int \cos^5(x) \sin(7x) &= \frac{1}{64i} \sum_{k=0}^5 {5 \choose k} \left[ \frac{e^{(2k+2)ix}}{(2k+2)i} - \frac{e^{(2k-12)ix}}{(2k-12)i} \right] \\ &= -\frac{1}{128} \sum_{k=0}^5 {5 \choose k} \left[ \frac{e^{(2k+2)ix}}{k+1} + \frac{e^{(2k-12)ix}}{6-k} \right]. \end{align*} Therefore, \begin{align*} \int_0^{\pi/2} \cos^5(x) \sin(7x) &= -\frac{1}{128} \sum_{k=0}^5 {5 \choose k} \left[ \frac{e^{(2k+2)i\pi/2} - 1}{k+1} + \frac{e^{(2k-12)i\pi/2} - 1}{6-k} \right] \\ &= -\frac{1}{128} \sum_{k=0}^5 {5 \choose k} \left[ \frac{(-1)^{k+1} - 1}{k+1} + \frac{(-1)^{k-6} - 1}{6-k} \right] \\ &= -\frac{1}{128} \left[ \frac{-2 {5 \choose 0}}{0 + 1} + \frac{-2 {5 \choose 2}}{2 + 1} + \frac{-2 {5 \choose 4}}{4 + 1} + \frac{-2 {5 \choose 1}}{6 - 1} + \frac{-2 {5 \choose 3}}{6 - 3} + \frac{-2 {5 \choose 5}}{6 - 5} \right] \\ &= \frac{4}{128} \left[ \frac{1}{1} + \frac{10}{3} + \frac{5}{5} \right] \\ &= \frac{1}{32} \cdot \frac{16}{3} \\ &= \boxed{\frac{1}{6}}. \end{align*}

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  • $\begingroup$ But thats $\cos(hx) , sin(hx)$ formula right , i have seen this exponential forms but we haven't been taught to use it anywhere yet , but your answer is very much self-explanatory , thanks for the answer , its always fun when we are able to solve big questions in different ways $\endgroup$ – Sujith Sizon Oct 12 '15 at 17:38
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$\int \cos^5x\sin7x dx = \int \cos^5x(\cos6x\sin x+\cos x\sin6x) dx = \int\cos6x\cos^5x\sin x+\cos^6x\sin6x dx$.

Let $u = \cos^6x$ and $v=\cos6x$ so that $du = -6\cos^5x\sin x dx$ and $dv = -6\sin6x$. So the original integral becomes: $-\frac{1}{6}\int udv + vdu$ And we get $-\frac{1}{6}\cos^6x\cos6x$. Applying limits then gives the answer of $\frac{1}{6}$.

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  • $\begingroup$ Thanks thats a really nice substitution $\endgroup$ – Sujith Sizon Oct 12 '15 at 17:23
  • $\begingroup$ But one negative sign will come outside right $\endgroup$ – Sujith Sizon Oct 12 '15 at 17:24
  • $\begingroup$ Oops yeh. I dropped my negative sign. $\endgroup$ – Ian Miller Oct 12 '15 at 17:27

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