Solutions1

# Solutions1 - sec 3 x tan x dx = Z sec 2 x sec x tan x dx =...

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MATH 2007B Test 1 Solutions October 3, 2011 [Marks] 1. Evaluate 2 x ( x 2 + 1) 23 dx [2] Solution: u = x 2 + 1 2 x ( x 2 + 1) 23 dx = u 23 du = u 24 24 + C = ( x 2 + 1) 24 24 + C . 2. Evaluate x 3 3 x 1 ( x 2 1) dx [4] Solution: u = x 3 3 x 1 x 3 3 x 1 ( x 2 1) dx = 1 3 x 3 3 x 1 (3 x 2 3) dx = 1 3 u du = 2 9 u 3 / 2 + C = 2 9 ( x 3 3 x 1) 3 / 2 + C . 3. Evaluate 4 xe 2 x dx [4] Solution: 4 xe 2 x dx = 2 xe 2 x 2 e 2 x dx = 2 xe 2 x e 2 x + C , integrating by parts. 4. Evaluate ( x 2 + 2) cos( x ) dx [4] Solution: ( x 2 + 2) cos( x ) dx = ( x 2 + 2) sin( x ) 2 x sin( x ) dx = ( x 2 + 2) sin( x ) + 2 x cos( x ) 2 cos( x ) dx = ( x 2 + 2) sin( x ) + 2 x cos( x ) 2 sin( x ) + C , integrating by parts twice. 5. Evaluate sin 5 ( x ) cos 3 ( x ) dx [4] Solution: sin 5 ( x ) cos 3 ( x ) dx = sin 5 ( x ) cos 2 ( x ) cos( x ) dx = sin 5 ( x )[1 sin 2 ( x )] cos( x ) dx = [sin 5 ( x ) sin 7 ( x )] cos( x ) dx = u 5 u 7 du = u 6 6 u 8 8 + C = sin 6 ( x ) 6 sin 8 ( x ) 8 + C , where u = sin( x ). 6. Evaluate tan 4 ( x ) sec 4 ( x ) dx [4] Solution: tan 4 ( x ) sec 4 ( x ) dx = tan 4 ( x ) sec 2 ( x ) sec 2 ( x ) dx = tan 4 ( x )[tan 2 ( x ) + 1] sec 2 ( x ) dx = [tan 6 ( x ) + tan 4 ( x )] sec 2 ( x ) dx = u 6 + u 4 du = u 7 7 + u 5 5 + C = tan 7 ( x ) 7 + tan 5 ( x ) 5 + C , where u = tan( x ).

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2 7. Evaluate sec 3 ( x ) tan( x
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Unformatted text preview: sec 3 ( x ) tan( x ) dx = Z sec 2 ( x ) sec( x ) tan( x ) dx = Z u 2 du = u 3 3 + C = sec 3 ( x ) 3 + C , where u = sec( x ). Alternatively, Z sec 3 ( x ) tan( x ) dx = Z sin( x ) cos 4 ( x ) dx = − Z 1 u 4 du = u-3 3 + C = 1 3 cos 3 ( x ) + C = sec 3 ( x ) 3 + C , where u = cos( x ). 8. Evaluate Z 4 cos 4 ( x ) dx [4] Solution: Z 4 cos 4 ( x ) dx = Z 4 ± 1 + cos(2 x ) 2 ² 2 dx = Z 1 + 2 cos(2 x ) + cos 2 (2 x ) dx = Z 1 + 2 cos(2 x ) + 1 + cos(4 x ) 2 dx = Z 3 2 + 2 cos(2 x ) + 1 2 cos(4 x ) dx = 3 2 x + sin(2 x ) + 1 8 sin(4 x ) + C ....
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