How do you evaluate the definite integral (sint+cost)dt from [0,π4]?

1 Answer
Nov 6, 2016

π40(sint+cost)dt=1

Explanation:

We have to know that:

  • (f(x)+g(x))dx=f(x)dx+g(x)dx
  • sintdt=cost+C
  • costdt=sint+C

Combining these and keeping the bounds of the integral, we see that:

π40(sint+cost)dt=[cost+sint]π40

Evaluating at t=π4 then subtracting the function evaluated at t=0:

=[cos(π4)+sin(π4)][cos0+sin0]

=(22+22)(1+0)

=0+1

=1