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]