ज्ञात करें `int x sin 2x cos 3x dx`

Explanation

`I=int" x sin 2x cos 3x dx "= 1/2 int" x 2 3x sin 2x dx"`
`1/2* int x *( sin 5x-sin x)dx = 1/2 int x sin 5x dx - 1/2 int x sin x dx" ...(1)"`
अब `int x sin 5x dx = x(-(cos 5x)/5)-int 1(-(cos 5x)/5)dx`
`=-(xcos 5x)/5 +1/5intcos 5x dx`
`=-(x cos 5x )5+1/4 int cos 5x dx`
`= -(x cos 5x)/5+1/5 (sin 5x)/5 = -(x cos 5x)/5 +(sin 5x)/25" ....(2)"`
पुनः, `int x sin x dx = x =- ( - cos x ) - int 1(- cosx ) dx`
`=-x cos x + int cas x dx = - cos x + sin x " ...(3)"`
(1) से, `I = 1/2 (-(x cos 5x)/5 + (sin 5x)/25)-1/2 ( -xcos x + sin x )+c`
`=-(x cos 5x)/10+(sin 5x)/50 +(x cos x)/2 -(sin x )/2+c`