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

Step-by-step explanation:

Given  function:

Now,

let be 'x'

Therefore,

= dx

or

= 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

the correct answer is 5

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