prob5sol - Math 151 Spring 2008 Problem Set # 5 Solutions...

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Math 151 Spring 2008 Problem Set # 5 Solutions Trigonometric and Hyperbolic Integrals and Substitutions 1. We set u =s in( x ) so that du =cos( x ) dx : Z cos 3 ( x )sin 2 ( x ) dx = Z cos 2 ( x 2 ( x )cos( x ) dx = Z ¡ 1 sin 2 ( x ) ¢ sin 2 ( x x ) dx = Z ¡ 1 u 2 ¢ u 2 du = Z ¡ u 2 u 4 ¢ du = 1 3 u 3 1 5 u 5 = 1 3 sin 3 ( x ) 1 5 sin 5 ( x ) . 2. u x ) so that du = sin ( x ) dx : Z sin 3 ( x )cos 4 ( x ) dx = Z sin 2 ( x 4 ( x )sin( x ) dx = Z ¡ 1 cos 2 ( x ) ¢ cos 4 ( x x ) dx = Z ¡ 1 u 2 ¢ u 4 du = Z ¡ u 4 u 6 ¢ du = 1 5 u 5 + 1 7 u 7 = 1 5 cos 5 ( x )+ 1 7 cos 7 ( x ) . 3. u inh( x ) so that du =cosh( x ) dx : Z cosh 3 ( x )sinh 2 ( x ) dx = Z cosh 2 ( x 2 ( x )cosh( x ) dx = Z ¡ sinh 2 ( x )+1 ¢ sinh 2 ( x x ) dx = Z ¡ u 2 +1 ¢ u 2 du = Z ¡ u 4 + u 2 ¢ du = 1 5 u 5 + 1 3 u 3 = 1 5 sinh 5 ( x 1 3 sinh 3 ( x ) . 1
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4. We set u =cosh( x ) so that du =sinh( x ) dx : Z sinh 3 ( x )cosh 4 ( x ) dx = Z sinh 2 ( x 4 ( x )sinh( x ) dx = Z ¡ cosh 2 ( x ) 1 ¢ cosh 4 ( x x ) dx = Z ¡ u 2 1 ¢ u 4 du = Z ¡ u 6 u 4 ¢ du = 1 7 u 7 1 5 u 5 = 1 7 cosh 7 ( x ) 1 5 cosh 5 ( x ) . 5. Z cos 2 ( x ) dx = Z 1 2 (1 cos (2 x )) dx = x 2 1 2 Z cos (2 x ) dx = x 2 1 2 μ 1 2 sin (2 x ) = x 2 1 4 sin (2 x ) μ = x 2 1 2 sin ( x )cos( x ) . 6. u x ) and dv x ) dx so that du x ) dx and v = Z cosh ( x ) dx x ) . Thus, Z cosh 2 ( x ) dx = Z cosh ( x )cosh( x ) dx = Z udv = uv Z vdu =c o s h ( x x ) Z sinh ( x x ) dx o s h ( x x ) Z sinh 2 ( x ) dx o s h ( x x ) Z ¡ cosh 2 ( x ) 1 ¢ dx o s h ( x x )+ x Z cosh 2 ( x ) dx. 2
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Therefore, 2 Z cosh 2 ( x )=cosh( x )sinh( x )+ x, so that Z cosh 2 ( x )= 1 2 cosh ( x x 1 2 x 7. We set u =s inh( x ) and dv =sinh( x ) dx so that du =cosh( x ) dx and v = Z sinh ( x ) dx x ) . Thus, Z sinh 2 ( x ) dx = Z sinh ( x x ) dx = Z udv = uv Z vdu i n h ( x )cosh( x ) Z cosh ( x x ) dx =c o s h ( x x ) Z cosh 2 ( x ) dx o s h ( x x ) Z ¡ sinh 2 ( x )+1 ¢ dx o s h ( x x ) x Z sinh 2 ( x ) dx. Therefore, 2 Z sinh 2 ( x x x ) x, so that Z sinh 2 ( x 1 2 cosh ( x x ) 1 2 x. 3
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8. We set x =3sin( u ) so that u =arcsin ³ x 3 ´ .dx =3cos( u ) du and p 9 x 2 u ) . Thus, Z p 9 x 2 dx = Z (3 cos ( u )) (3 cos ( u )) du =9 Z cos 2 ( u ) du μ 1 2 cos ( u )sin( u )+ 1 2 u = 9 2 cos ( u u 9 2 u = 9 2 à 9 x 2 3 ! ³ x 3 ´ + 9 2 arcsin ³ x 3 ´ = 1 2 x p 9 x 2 + 9 2 arcsin ³ x 3 ´ . 9. We have x 2 4 + y 2 9 =1 y 2 9 x 2 4 y 2 μ 1 x 2 4 Therefore, the upper part of the ellipse is the graph of y =3 r 1 x 2 4 = 3 2 p 4 x 2 , 2 x 2 . By symmetry, the area of the region inside the ellipse is Z 2 2 3 2 p 4 x 2 dx = 3 2 Z 2 2 p 4 x 2 dx x =2sin( u ) so that u ³ x 2 ´ ,dx =2cos( u ) du and p 4 x 2 u ) . It is practical to implemen t the de f nite integral version of the substitution rule: 3 2 Z 2 2 p 4 x 2 dx = 3 2 Z arcsin(1) arcsin( 1) (2 cos ( u )) (2 cos ( u )) du =6 Z π/ 2 π/ 2 cos 2 ( u ) du à 1 2 cos ( u u 1 2 u ¯ ¯ ¯ ¯ π/ 2 π/ 2 ! μ 1 2 cos ³ π 2 ´ sin ³ π 2 ´ + 1 2 ³ π 2 ´ 6 μ 1 2 cos ³ π 2 ´ sin ³ π 2 ´ + 1 2 ³ π 2 ´ π 4
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10.
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prob5sol - Math 151 Spring 2008 Problem Set # 5 Solutions...

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