Evaluate the integral Integral from nothing to nothing ∫ StartFraction 3 Over t Superscript 4 EndFraction 3 t4 sine left parenthesis StartFraction 1 Over t cubed EndFraction minus 6 right parenthesis sin 1 t3 −6dt

Respuesta :

Answer:

[tex]\cos (\frac{1}{t^3}-6)} + c[/tex]

Step-by-step explanation:

Given  function:

[tex]\int {\frac{3}{t^4}\sin (\frac{1}{t^3}-6)} \, dt[/tex]

Now,

let [tex]\frac{1}{t^3}-6[/tex] be 'x'

Therefore,

[tex]d(\frac{1}{t^3}-6)[/tex] = dx

or

[tex]\frac{-3}{t^4}dt[/tex] = dx

on substituting the above values in the equation, we get

⇒ ∫ - sin (x) . dx

or

cos (x) + c                      [ ∵ ∫sin (x) . dx = - cos (x)]

Here,

c is the integral constant

on substituting the value of 'x' in the equation, we get

[tex]\cos (\frac{1}{t^3}-6)} + c[/tex]