00:01
This problem is from chapter 7, section 2, problem number 14 in the book calculus, early transcendental's, 8th edition by james stewart.
00:11
Here we have an indefinite integral of sine squared of 1 over t, all divided by t squared.
00:18
So before we start to integrate here, we can apply a u substitution.
00:22
Let's take u to be 1 over t so that du using the power rule from differential calculus, is negative 1 over t squared d t.
00:38
Equivalently negative d u is 1 over t squared d t so if we apply our u substitution here we can rewrite our original integral as negative integral sine squared of u d u now we can apply an identity for sign squared using a trigonometric identity let's rewrite this expression.
01:17
Here, the integrand.
01:22
We have integral, negative integral, and we can rewrite the sign squared as 1 minus cosine of 2 u, all divided by 2...