271_HW7.2 - MAT 271 Summer 2011 Calculus II Scott Zinzer...

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MAT 271, Summer 2011 Calculus II, Scott Zinzer 7.2 HW 7.2. Trigonometric Integrals. 16. Evaluate the integral Z sin - 3 / 2 x cos 3 x dx Solution: We save a single factor of cos x and use the pythagorean identity to obtain Z sin - 3 / 2 x cos 3 x dx = Z sin - 3 / 2 x cos 2 x (cos x dx ) = Z sin - 3 / 2 x (1 - sin 2 x )(cos x dx ) Now we make the substitution u = sin x . Then du = cos x dx and we have Z sin - 3 / 2 x (1 - sin 2 x )(cos x dx ) = Z u - 3 / 2 (1 - u 2 ) du = Z u - 3 / 2 - u 1 / 2 du = - 2 u - 1 / 2 - 2 3 u 3 / 2 + C = - 2 sin x - 2 3 sin 3 / 2 x + C 28. Evaluate the integral Z csc 10 x cot 3 x dx Solution: We save a factor of csc 2 x and use the pythagorean identity to obtain Z csc 10 x cot 3 x dx = Z (csc 2 x ) 4 cot 3 x (csc 2 x dx ) = Z (cot 2 x + 1) 4 cot 3 x (csc 2 x dx ) Now we make the substitution u = cot x , so that du = - csc 2 x dx . This gives Z (cot 2 x + 1) 4 cot 3 x (csc 2 x dx ) = - Z ( u 2 + 1) 4 u 3 du = - Z ( u 8 + 4 u 6 + 6 u 4 + 4 u 2 + 1 ) u 3 du = - Z u 11 + 4 u 9 + 6 u 7 + 4 u 5 + u 3 du = - 1 12 u 12 - 2 5 u 10 - 3 4 u 8 - 2 3 u 6 - 1 4 u 4 + C = - 1 12 cot 12 x - 2 5 cot 10 x - 3 4 cot 8 x - 2 3 cot 6 x - 1 4 cot 4 x + C
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MAT 271, Summer 2011 Calculus II, Scott Zinzer 7.2 HW 30. Evaluate the integral Z tan 5 θ sec 4 θ dθ Solution: We save a factor of sec 2 θ and use the pythagorean identity to obtain Z tan 5 θ sec 4 θ dθ = Z tan 5 θ sec 2 θ (sec 2 θ dθ ) = Z tan 5 θ (1 + tan 2 θ )(sec
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Unformatted text preview: u = tan θ , so that du = sec 2 θ dθ . This gives Z tan 5 θ (1 + tan 2 θ )(sec 2 θ dθ ) = Z u 5 (1 + u 2 ) du = Z u 5 + u 7 du = 1 6 u 6 + 1 8 u 8 + C = 1 6 tan 6 θ + 1 8 tan 8 θ + C 44. Evaluate the integral Z π/ 4-π/ 4 tan 3 x sec 2 xdx Solution: Recall that sin x is an odd function and cos x is an even function, i.e. sin(-x ) =-sin x and cos(-x ) = cos x . This means that tan(-x ) = sin(-x ) cos(-x ) =-sin x cos x =-tan x , i.e. that tan x is an odd function and sec(-x ) = 1 cos(-x ) = 1 cos x = sec x , i.e. that sec x is an even function. Therefore tan 3 (-x ) sec 2 (-x ) = (tan(-x )) 3 (sec(-x )) 2 = (-tan x ) 3 (sec x ) 2 =-tan 3 x sec 2 x so that tan 3 x sec 2 x is an odd function. Therefore Z π/ 4-π/ 4 tan 3 x sec 2 xdx = 0...
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