1. [40.15 Points] DETAILS SCALCET9 7.3.AE.OO1. 01100 Submissions Used MY NOTES ASK YOUR TEACHER Example 1 Video Example 41)) [V 4 x2 dx. x2 Evaluate Solution Letx = 2 5mm), where :r s 6' s :. Then dx = |:' GB and 2 2 V4 x2 = \\/4 4 sin2(9 = \\/4 0052(9) = 2|cos(6)| = 2 cos(9). (Note that cos(9) a 0 because a s 9 s 1.) Thus, the inverse substitution rule gives 2 T fW wrZ (S )de X2 4 sinzw) c0520?) =1 : r = [cotzm d8 _ [(CSCZW) 1) (16' Since this is an indenite integral, we must return to the original variable x. This can be done either by using trigonometric identities to express cot(6') in terms of sin(6') = g or by drawing a diagram' as in the gure below, where 3 is interpreted as an angle of a right triangle. v4-12 (D Since 5in(l9) = %, we label the opposite side and the hypotenuse as having lengths x and 2. Then the Pythagorean theorem gives the length of the adjacent side as \\f 4 _ x2, so we can simply read the value of cot(6') from the figure. W) = S X (Although 6 > D in the diagram, this expression for cot(l9) is valid even when 6