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Four. 97909_07_ch07_p474-483. qk:97909_07_ch07_p474-483 482 bankruptcy 7 9/21/10 5:28 PM web page 482 options OF INTEGRATION notice As instance five illustrates, hyperbolic substitutions can be utilized instead of trigonometric substitutions and infrequently they result in easier solutions. yet we frequently use trigonometric substitutions simply because trigonometric identities are extra favourite than hyperbolic identities. As instance 6 indicates, trigonometric substitution is typically a good suggestion while ͑x 2 ϩ a 2 ͒ n͞2 happens in an fundamental, the place n is any integer. an identical is correct while ͑a 2 Ϫ x 2 ͒ n͞2 or ͑x 2 Ϫ a 2 ͒ n͞2 happen. instance 6 locate y three s3͞2 x3 dx. ͑4x ϩ 9͒3͞2 2 zero answer First we word that ͑4x 2 ϩ 9͒3͞2 ෇ ͑s4x 2 ϩ nine )3 so trigonometric substitution is suitable. even though s4x ϩ nine isn't relatively one of many expressions within the desk of trigonometric substitutions, it turns into one in all them if we make the initial substitution u ෇ 2x. after we mix this with the tangent substitution, we have now x ෇ 32 tan ␪, which provides dx ෇ 32 sec 2␪ d␪ and a couple of s4x 2 ϩ nine ෇ s9 tan 2␪ ϩ nine ෇ three sec ␪ whilst x ෇ zero, tan ␪ ෇ zero, so ␪ ෇ zero; whilst x ෇ 3s3͞2, tan ␪ ෇ s3 , so ␪ ෇ ␲͞3. y three s3͞2 zero 27 three x3 ␲͞3 eight tan ␪ dx ෇ y zero ͑4x 2 ϩ 9͒3͞2 27 sec3␪ ෇ 163 y ␲͞3 zero ෇ 163 y ␲͞3 zero three 2 sec 2␪ d␪ three tan three␪ ␲͞3 sin ␪ d␪ ෇ 163 y d␪ zero sec ␪ cos2␪ 1 Ϫ cos 2␪ sin ␪ d␪ cos 2␪ Now we replacement u ෇ cos ␪ in order that du ෇ Ϫsin ␪ d␪. while ␪ ෇ zero, u ෇ 1; whilst ␪ ෇ ␲͞3, u ෇ 12. for this reason y three s3͞2 zero 2 x3 1͞2 1 Ϫ u three dx ෇ Ϫ du sixteen y 1 ͑4x 2 ϩ 9͒3͞2 u2 ෇ 163 y 1͞2 1 ෇ 163 instance 7 review y [( 1 2 ͫ ͬ ͑1 Ϫ u Ϫ2 ͒ du ෇ 163 u ϩ ] ϩ 2) Ϫ ͑1 ϩ 1͒ ෇ 1 u 1͞2 1 three 32 x dx. s3 Ϫ 2x Ϫ x 2 answer we will be able to remodel the integrand right into a functionality for which trigonometric substi- tution is acceptable via first finishing the sq. less than the basis signal: three Ϫ 2x Ϫ x 2 ෇ three Ϫ ͑x 2 ϩ 2x͒ ෇ three ϩ 1 Ϫ ͑x 2 ϩ 2x ϩ 1͒ ෇ four Ϫ ͑x ϩ 1͒2 this implies that we make the substitution u ෇ x ϩ 1. Then du ෇ dx and x ෇ u Ϫ 1, so y x uϪ1 dx ෇ y du 2 s3 Ϫ 2x Ϫ x s4 Ϫ u 2 97909_07_ch07_p474-483. qk:97909_07_ch07_p474-483 9/21/10 5:30 PM web page 483 part 7. three TRIGONOMETRIC SUBSTITUTION 483 We now alternative u ෇ 2 sin ␪, giving du ෇ 2 cos ␪ d␪ and s4 Ϫ u 2 ෇ 2 cos ␪, so determine five indicates the graphs of the integrand in instance 7 and its indefinite crucial (with C ෇ 0). that is which? x y s3 Ϫ 2x Ϫ x three 2 2 sin ␪ Ϫ 1 2 cos ␪ d␪ 2 cos ␪ dx ෇ y ෇ y ͑2 sin ␪ Ϫ 1͒ d␪ 2 _4 ෇ Ϫ2 cos ␪ Ϫ ␪ ϩ C ͩͪ ෇ Ϫs4 Ϫ u 2 Ϫ sinϪ1 _5 1–3 overview the quintessential utilizing the indicated trigonometric substitution. cartoon and label the linked correct triangle. three. ͩ ͪ xϩ1 2 21. y zero. 6 zero x2 dx s9 Ϫ 25x 2 y dx x s4 Ϫ x 2 x ෇ 2 sin ␪ 23. y s5 ϩ 4x Ϫ x y x3 dx 2 ϩ four sx x ෇ 2 tan ␪ 25. y y sx 2 Ϫ four dx x x ෇ 2 sec ␪ 27. y sx 29. y x s1 Ϫ x 2 y 22. 1 sx 2 ϩ 1 dx zero dt st 2 Ϫ 6t ϩ thirteen 24. y x dx sx ϩ x ϩ 1 26. y ͑3 ϩ 4x Ϫ 4x ϩ 2x dx 28. y ͑x 30. y 2 dx 2 2 four dx x2 2 ␲ ͞2 zero 4–30 assessment the critical. four. y 1 zero five. y x 3s1 Ϫ x 2 dx 1 2 s2 7.