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I'm looking for a quick and elegant way to calculate the integral: $$\int x\sin ax \cos x\, dx$$

It's doable by using $$\cos x=\frac{e^{ix}+e^{-ix}}2, ~~\sin ax=\frac{e^{iax}-e^{-iax}}{2i},$$ and then integrating by parts, but I find that solution too long and brute-force. Can you solve it alternatively?

Edit: $$\sin ax\cos x=\frac{\sin (a+1)+\sin (a-1)x}2$$ does the trick.

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Use Werner formula:

$$2\sin A\cos B=\sin(A+B)+\sin(A-B)$$

Now integrate by parts $$\int x\sin(px)\ dx=x\int\sin(px)\ dx-\int\left(\dfrac{dx}{dx}\int\sin(px)\ dx\right)dx$$

OR as $$\dfrac{d(x\cos(px))}{dx}=\cos(px)-px\sin(px)$$

Integrate both sides, $$x\cos(px)+K=\int\cos(px)\ dx-p\int x\sin(px)\ dx$$

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Just with some (mild) trigonometry to linearise : $$\sin ax\,\cos x=\frac12\bigl(\sin(a+1)x+\sin(a-1)x\bigr),$$ then integration by parts to remove the $x$ factor.

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If you don't have trig identities at your fingertips, note that

$$\int x\sin ax\cos x\,dx=-{dI\over da}$$

where

$$I(a)=\int\cos ax\cos x\,dx$$

Integration by parts twice tells us

$$\begin{align} I(a) &=\cos ax\sin x+a\int\sin ax\sin x\,dx\\ &=\cos ax\sin x-a\sin ax\cos x+a^2\int\cos ax\cos x\,dx\\ &=\cos ax\sin x-a\sin ax\cos x+a^2I(a) \end{align}$$

hence

$$I(a)={\cos ax\sin x-a\sin ax\cos x\over1-a^2}$$

and thus

$$\begin{align} \int x\sin ax\cos x\,dx &={(1-a^2)(-x\cos ax\sin x-\sin ax\cos x-ax\cos ax\cos x)+2a(\cos ax\sin x-a\sin ax\cos x)\over(1-a^2)^2}\\ &={(a^2x-x+2a)\cos ax\sin x-(a^2+1)\sin ax\cos x+a(a^2-1)\cos ax\cos x\over(1-a^2)^2} \end{align}$$

Remark: At the end of all this, one should add the obligatory "constant" of integration $C$, which in this case is a function, $C(a)$. Among other things, the constant of integration can help reconcile two different-looking answers obtained by two different approaches to evaluating the integral. (Its main purpose is to ensure you don't have points taken off by a picky instructor....)

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$$2\int x\sin ax\cos x dx=\int x\sin(a-1)x dx+\int x\sin(a+1)xdx =I_1+I_2$$ by the formula: $2\sin A\cos B=\sin(A+B)+\sin(A-B)$

Consider $I_1:$ set $u=(a-1)x$, then $du=(a-1)dx$, then

$$\int x\sin(a-1)x dx=\int \frac{u\sin u}{(a-1)^2}du=\frac{1}{(a-1)^2}\int u\sin u du=\underbrace{\cdots}_{\text{integration by parts}}=\frac{1}{(a-1)^2}\big(\sin u- u\cos u\big)$$

Consider now $I_2:$ set $v=(a+1)x$, then $dv=(a+1)dx$, then

$$\int x\sin(a+1)x dx=\int \frac{v\sin v}{(a+1)^2}dv=\frac{1}{(a+1)^2}\int v\sin v dv=\underbrace{\cdots}_{\text{integration by parts}}=\frac{1}{(a+1)^2}\big(\sin v- v\cos v\big)$$

Summing up $I_1+I_2$ we get $$ \frac{1}{(a-1)^2}\big(\sin u- u\cos u\big)+\frac{1}{(a+1)^2}\big(\sin v- v\cos v\big)$$

Finally: $$\int x\sin ax\cos x dx=\frac12\Big[\frac{1}{(a-1)^2}\big(\sin (a-1)x- (a-1)x\cos (a-1)x\big)+ \frac{1}{(a+1)^2}\big(\sin (a+1)x- (a+1)x\cos (a+1)x\big)\Big]$$

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  • $\begingroup$ DIfferentiating your result doesn't yield the integrand. I'm not sure how you can sum up $I_{1}$ and $I_{2}$ when they have different arguments. $\endgroup$ – Mattos Apr 20 '18 at 12:38
  • $\begingroup$ I use the same variable for both and I made a mistake! I edit my post, thanks for your comment :) $\endgroup$ – InsideOut Apr 20 '18 at 13:04

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