find integral ∫0π2sin⁡(40x)sin⁡5xdx My attempt -writing sin⁡40x=sin⁡40x−sin⁡30x+sin⁡30x−sin⁡20x+sin⁡20x−sin⁡10x+sin⁡10x

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find integral ∫0π2sin⁡(40x)sin⁡5xdx  My attempt -writing  sin⁡40x=sin⁡40x−sin⁡30x+sin⁡30x−sin⁡20x+sin⁡20x−sin⁡10x+sin⁡10x

Toni Scott · Accepted Answer

Hint: Note that  sin⁡40x−sin⁡10xsin⁡5x=2(cos⁡15x+cos⁡25x+cos⁡35x)

temnimam2 · Accepted Answer

You can generalize to any integer:  sin⁡(nx)sin⁡x=e∈x−e−∈xeix−e−ix=e−i(n−1)x(1−ei2nx1−ei2x)=e−(n−1)x∑k=0n−1ei2kx=∑k=0n−1ei(2k+1−n)x  From there you have two possibilities, notice the indexes (2k+1-n) take values  {−(n−1)−(n−3)…−11…(n−3)(n−1) n even−(n−1)−(n−3)…−202…(n−3)(n−1)n odd  Therefore you can regroup terms two by two eimx+e−imx=2cos⁡(mx) and end up with a sum of cos⁡(mx) where m varies with a step 2  These cosinus are easy to integrate and will give you a rational value for the integral. Notice that for n odd, you have also a constant term that will make a π2 value appear.   The second possibility is to integrate the exponential directly  Then In=∑k=0n−1eiπ2(2k+1−n)−1(2k+1−n)i  Notice also that ∫05π2sin⁡8xsin⁡xdx=∫0π2sin⁡8xsin⁡xdx since the integral on [0,2π] is zero.   So overall our integral is just 15I8

Vasquez · Accepted Answer

Expand the numerator, then divide each term by sin⁡5x (or sin⁡y after substituting y=5x ) I=∫0π2sin⁡40xsin⁡5xdx =15∫05π2sin⁡8ysin⁡ydy =15∫05π2f(y)sin⁡ydy where f(y)=Im(e8iy) =8cos7⁡(y)sin⁡(y)−56cos5⁡ysin3⁡y+56cos3⁡ysin5⁡y−8cos⁡ysin7⁡y ⇒I=15∫05π2(8cos7⁡y−56cos5⁡ysin2⁡y+56cos3⁡ysin4⁡y−8cos⁡ysin6⁡y)dy

find integral \int_{0}^{\frac{\pi}{2}} \frac{\sin(40x)}{\sin 5x}dx My attempt -writing \sin 40x=\sin 40x-\sin 30x

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