Question

If u and v are two differentiable functions of x, then prove that `intu*v*dx = u*intv  dx - int(d/dx u)(intv  dx)dx`. Hence evaluate: `intx cos x  dx`

Answer 1

Let `intv  dx` = w

`\implies` v = `(dw)/dx`   ...(i)

Consider, `d/dx(u*w) = u*d/dxw + w*d/dx u`

= `u*v + w*(du)/dx`   ...[From (i)]

By definition of integration

u.w. = `int[uv + w(du)/dx]dx`

= `intuv  dx + int(du)/dx*w  dx`

∴ `u*intv  dx = intuv  dx + int[(du)/dx intvdx]dx`   ...[From (i)]

∴ `intuv  dx = uintv  dx - int((du)/dx)(intv  dx)dx`

Now, I = `intx cos x  dx`,

Here u = x,

v = cos x

∴ I = `x int cosx  dx - int(d/dx x)(intcosx  dx)dx`

= `xsinx - int(1) sinx  dx`

= x sin x + cos x + c