# PW8Solutions - MATH 1700 Problem Workshop 8 Solutions 1 Let...

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MATH 1700 Problem Workshop 8 Solutions 1. Let w = cos x dw = - sin xdx. I = Z sin 5 x cos 2 xdx = - Z sin 4 x cos 2 xdw = - Z (1 - w 2 ) 2 w 2 dw = - Z ( w 6 - 2 w 4 + w 2 ) dw = - w 7 7 + 2 w 5 5 - w 3 3 + C = - cos 7 x 7 + 2 cos 5 x 5 - cos 3 x 3 + C 2. Using sin 2 x = 1 - cos 2 x 2 and cos 2 x = 1 + cos 2 x 2 we get I = Z 1 - cos 2 x 2 2 dx = Z 1 4 - cos 2 x 2 + cos 2 x 4 dx = Z 1 4 - cos 2 x 2 + 1+cos 4 x 2 4 dx = Z 3 8 - cos 2 x 2 + cos 4 x 8 dx = 3 8 x - sin 2 x 4 + sin 4 x 32 + C 3. Let w = csc x dw = - csc x cot xdx. Using cot 2 x = csc 2 x - 1 , we get 1
I = - Z cot 4 x csc 2 xdw = - Z ( w 2 - 1) w 2 dw = - Z ( w 6 - 2 w 4 + w 2 ) dw = - w 7 7 + 2 w 5 5 - w 3 3 + C = - csc 7 x 7 + 2 csc 5 x 5 - csc 3 x 3 + C 4. The integral can be rearranged to get I = Z 1 sin x cos 2 x dx = Z cot 2 x csc x dx = Z csc 2 x - 1 csc x dx = Z csc x - 1 csc x dx = Z (csc x - sin x ) dx = - ln | csc x + cot x | + cos x + C 5. Using x = 2 sin θ dx = 2 cos θdθ. 2
I = Z 1 x 4 - x 2 dx = Z 2 cos θdθ 2 sin θ p 4 - (2 sin θ ) 2 = Z 2 cos θdθ 2 sin θ p 4 - 4 sin 2 θ = Z 2 cos θdθ 2 sin θ 4 cos 2 θ = Z 2 cos θdθ 4 sin θ cos θ = 1 2 Z csc θdθ = - 1 2 ln | csc θ + cot θ | + C Using sin θ = x 2 we could use triangles to get csc θ = 2 x , cot θ = 4 - x 2 x .