What is $\int_0^{\pi/2}\cos^{10}x (\sin 12x )\,dx\;$?

Question

$$\int_0^{\pi/2}(\cos^{10}x )(\sin 12x )dx$$

I am unable to figure out the trick this question expects me to use. Its not a simple by parts question.

If I denotes the integral then I is also equal to $\int_0 ^{\pi/2}(\sin^{10}x) (\cos 12 x)dx$

Adding the two I's , doesn't yield anything special.

How do I go about solving this problem? What's the trick to immediately recognize the method the integration questions expect me to use?

Answer 1

Hint: the integrand can be written as $$\begin{align} & ={{\cos }^{10}}\left( x \right)\sin \left( 11x+x \right) \\ & ={{\cos }^{10}}\left( x \right)\left( \sin \left( 11x \right)\cos \left( x \right)+\cos \left( 11x \right)\sin \left( x \right) \right) \\ & ={{\cos }^{11}}\left( x \right)\sin \left( 11x \right)+{{\cos }^{10}}\left( x \right)\cos \left( 11x \right)\sin \left( x \right) \\ & =-\frac{1}{11}\frac{d}{dx}\left[ {{\cos }^{11}}\left( x \right)\cos \left( 11x \right) \right] \\ \end{align}$$

Answer 2

I'd suggest rewriting this with exponentials using Euler's formula; then you just need to multiply it out and trivially integrate the exponentials. It's perhaps not the most elegant way to do it, but it's straightforward. The binomial theorem does most of the work of multiplying it out for you.

Answer 3

Hint: $$ \begin{align} \sin x\sin y &= \dfrac12(\cos(y - x) - \cos(y + x)) \\ \cos x\cos y &= \dfrac12(\cos(y + x) - \cos(y - x)) \\ \sin x\cos y &= \dfrac12(\sin(y + x) - \sin(y - x)) \end{align}$$

Also, $$\begin{align} \cos^{10}x\sin 12x &= (\cos^2x)^5\sin 12x \\ &= \left(\dfrac12(\cos 2x + 1)\right)^5\sin 12x \\ &= \dfrac1{32}\left(\cos^52x + 5\cos^42x + 10\cos^32x + 10\cos^2x + 5\cos 2x + 1\right)\sin 12x \end{align}$$ and so on. Using the product-to-sum formulas above and linearity, you should be able to solve the integral.

Answer 4

The solution holds in a word: linearise the expression, starting from Euler's formulæ. Here's how to start: \begin{align} \cos^{10}x\,\sin 12 x&=\Bigl(\frac{\mathrm e^{ix}+\mathrm e^{-ix}}{2}\Bigr)^{\!10}\,\frac{\mathrm e^{12ix}+\mathrm e^{-12ix}}{2i}\\ &=\frac{\mathrm e^{10ix}+10\mathrm e^{8ix}+45\mathrm e^{6ix}+\dotsm}{2^{10}}\cdot\frac{\mathrm e^{12ix}+\mathrm e^{-12ix}}{2i} \end{align} Can you proceed?

Answer 5

Hint : \begin{eqnarray*} \frac{d}{dx}(\cos^{11}x \cos(11x)) =-11\cos^{10} x \sin x \cos(11 x) - 11 \cos^{11} x \sin(11x). \end{eqnarray*} Now use the $\sin$ addition formula.

Answer 6

$$I = \int \cos^{10}x(\sin 12 x)dx$$

$$I = \Im\int \bigg[e^{i(12 x)}\cdot \bigg(\frac{e^{ix}+e^{-ix}}{2}\bigg)^{10}\bigg]dx$$

$$=\frac{1}{2^{10}}\Im\int \bigg[e^{i(2x)}\cdot \bigg(e^{i(2x)}+1\bigg)^{10}\bigg]dx$$

Now put $e^{i(2x)}+1=t,$ Then $\displaystyle 2ie^{i(2x)}dx = dt\Rightarrow e^{i(2x)}dx = \frac{dt}{2i}$

So $$=\Im\bigg[\frac{1}{2^{10}i}\int t^{10}dt\bigg] = \frac{1}{2^{11}\cdot 11}\Im\bigg[\bigg(\frac{e^{2ix}+1}{e^{ix}}\bigg)^{11}\cdot e^{11ix}\bigg]+\mathcal{C}$$

$$=-\frac{1}{11}\Im\bigg[\cos^{11}(x)\cdot \bigg(\cos 11 x+i\sin 11 x\bigg)\bigg]+\mathcal{C}$$

So $$I = -\frac{1}{11}\cos^{11}(x)\cdot \cos (11x)+\mathcal{C}$$

So $$\int^{\frac{\pi}{2}}_{0}\cos^{10}x(\sin 12 x)dx = -\frac{1}{11}\bigg[\cos^{11}(x)\cdot \cos(11)\bigg]\Bigg|^{\frac{\pi}{2}}_{0}=\frac{1}{11}.$$