Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 92. A major complaint of professors teaching calculus is that students don't have the appropriate background to work through the calculus course successfully. This text is targeted directly at this underprepared audience. This is a single-variable (2-semester) calculus text that incorporates a conceptual re-introduction to key precalculus ideas throughout the exposition as appropriate. This is the ideal resource for those schools dealing with poorly prepared students or for schools introducing a slower paced, integrated precalculus/calculus course | Trigonometric Integrals and Trigonometric Substitution 891 1 V9 1 V9 x2 1 1 1 3 x2 C arcsin dx Let u du dx 2 3 3 3 Method 2 We d love to get rid of that square root in the denominator. To do this let s replace 9 x2 by a perfect square. We can exploit the trigonometric identity 1 sin2 9 cos2 9 or 9 9 sin2 9 9 cos2 9. Let x 3 sin 9 where 9 e n n 2 2 dx 3 cos 9d9. Then V9 x2 9 9 sin2 9 V9 cos2 9 3Vcos2 9 i 1 dx i --------3 cos 9d91 J V9 x2 Wcos2 9 1 d9 9 C But x 3 sin 9 so 3 sin 9 and 9 arcsin 3 . 1 dx arcsin x C V9 x2 3 This second method will come in very handy particularly when the first method cannot be applied. Notice that although the integrand involves no trigonometry the antiderivative does. EXAMPLE Find 0 V4 x2 dx. SOLUTION The strategy is to replace 4 x2 by a perfect square in order to eliminate the square root. Let x 2 sin 9 where 9 e n n 2 dx 2 cos 9 d9. 4 x2 J4 4 sin2 9 y 4 cos2 9 2 cos2 9 What are the new endpoints of integration When x 0 When x 1 0 2 sin 9 sin 9 0 9 arcsin 0 0. 1 2 sin 9 sin 9 2 9 arcsin 6. 1Vcos2 9 is actually cos 91 as opposed to cos 9. In the context of this problem however Vcos2 9 can be replaced by cos 9 m in n 3 Imn L tn I n n I rl n 3 . tT rx n 1-L in i n m 1 because 9 nas Deen restricted to 2 and cos 9 0 on this interval. 892 CHAPTER 29 Computing Integrals 1 I 4 - x2 dx 2 cos2 0 2 cos 0 d0 0 0 I cos2 0 d0 Use cos2 0 - 1 cos 20 . 02 z. I 6 1 cos 20 d0 0 2 0 1 sin 20 ZL 6 0 2 n 1 sin n - o 2 3 1 V3 3 2 Clarifications and Justifications Notice that when we let x 2 sin 0 we are restricting x to lie between -2 and 2. That s fine because V4 x2 only makes sense for x in this interval. In both of the previous examples we ve said Vcos2 0 cos 0. Actually Vcos2 0 cos 01 which is equal to cos 0 only if cos 0 is nonnegative. It is because we ve restricted 0 to y y that we can write cos 01 cos 0. In fact the substitution x g 0 is only valid if g 0 is 1-to-1. In other words when we say x 2 sin 0 we must restrict 0. In this case
