7 sin 3 x cos 2 x dx solution Because the integrand is a combination of sine

7 sin 3 x cos 2 x dx solution because the integrand

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7. sin 3 x cos 2 x dx solution Because the integrand is a combination of sine and cosine functions with sine raised to an odd power, the substitution u = cos x is the most appropriate method for this integral. For each of the following, find the formula in the integral table at the back of the book that can be applied to find the integral: 8. 3 x 2 dx 5 x + 2 solution First write 3 x 2 dx 5 x + 2 = 3 x 2 dx 5 x + 2 . The integral on the right-hand side can then be evaluated using formula 94, u 2 a + bu du = 1 2 b 3 [ (a + bu) 2 4 a(a + bu) + 2 a 2 ln | a + bu |] + C, with a = 2 and b = 5. 9. 25 + 16 x 2 x 2 dx solution First let u = 4 x . Then du = 4 dx and 25 + 16 x 2 x 2 dx = 4 25 + u 2 u 2 du. The integral on the right-hand side can then be evaluated using formula 87, a 2 + u 2 u 2 du = − a 2 + u 2 u + ln u + a 2 + u 2 + C, with a = 5. 10. sec 3 ( 4 x) dx solution First let u = 4 x . Then du = 4 dx and sec 3 ( 4 x) dx = 1 4 sec 3 u du. For the integral on the right-hand side, formula 48, sec n u du = 1 n 1 tan u sec n 2 u + n 2 n 1 sec n 2 u du, with n = 3, can be applied. 11. x 2 x 2 + 2 x + 5 dx solution Firstcompletethesquareundertheradical, x 2 + 2 x + 5 = (x + 1 ) 2 + 4,andmakethesubstiution u = x + 1. Then du = dx , x = u 1, and x 2 x 2 + 2 x + 5 dx = (u 1 ) 2 u 2 + 4 du = u 2 u 2 + 4 du 2 u u 2 + 4 du + 1 u 2 + 4 du. For the first integral on the right-hand side, apply formula 89, u 2 du a 2 + u 2 = u 2 a 2 + u 2 a 2 2 ln u + a 2 + u 2 + C,
1018 C H A P T E R 8 TECHNIQUES OF INTEGRATION with a = 2, while for the third integral on the right-hand side, apply formula 88, du a 2 + u 2 = ln u + a 2 + u 2 + C, also with a = 2. For the second integral on the right-hand side, let w = a 2 + u 2 . Then dw = 2 u du , and 2 u u 2 + 4 du = dw w = w 1 / 2 dw, which can be evaluated using formula 1, u n du = u n + 1 n + 1 + C, with n = − 1 / 2. Exercises In Exercises 1–10, indicate a good method for evaluating the integral (but do not evaluate). Your choices are: substitution (specify u and du ), Integration by Parts (specify u and dv ), a trigonometric method, or trigonometric substitution (specify). If it appears that these techniques are not sufficient, state this. 1. x dx 12 6 x x 2 solution First complete the square under the radical, 12 6 x x 2 = 12 (x 2 + 6 x) = 12 + 9 (x 2 + 6 x + 9 ) = 21 (x + 3 ) 2 , and make the substitution u = x + 3. Then du = dx , x = u 3, and x dx 12 6 x x 2 = u 3 21 u 2 du = u 21 u 2 du 3 1 21 u 2 du. For the first integral on the right-hand side, make the substitution w = 21 u 2 ; for the second integral on the right-hand side, recognize the formula for the inverse sine or use the trigonometric substitution u = 21 sin θ . 2. 4 x 2 1 dx solution Use trigonometric substitution, with x = 1 2 sec θ . 3. sin 3 x cos 3 x dx solution Use one of the following trigonometric methods: rewrite sin 3 x = ( 1 cos 2 x) sin x and let u = cos x , or rewrite cos 3 x = ( 1 sin 2 x) cos x and let u = sin x . 4. x sec 2 x dx solution Use Integration by Parts, with u = x and dv = sec 2 x dx .

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