2000 Paper 3 Q2

Year: 2000
Paper: 3
Question Number: 2

Course: LFM Pure
Section: Integration

Difficulty: 1700.0 Banger: 1484.2

integration trigonometric substitution definite integral square root generalisation trigonometric identities

Problem

Use the substitution $x = 2-\cos \theta $ to evaluate the integral $$ \int_{3/2}^2 \left(x - 1 \over 3 - x\right)^{\!\frac12}\! \d x. $$ Show that, for $a < b$, $$ \int_p^q \left( x - a \over b - x\right)^{\!\frac12} \!\d x = \frac{(b-a)(\pi +3{\surd3} -6)}{12}, $$ where $p= {(3a+b)/4}$ and $q={(a+b)/2}$.

No solution available for this problem.

Rating Information

Difficulty Rating: 1700.0

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Banger Rating: 1484.2

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Problem source

Use the substitution   
$x = 2-\cos \theta $   
to evaluate the integral 
 
$$ 
\int_{3/2}^2 \left(x -  1 \over 3 - x\right)^{\!\frac12}\! \d x.  
$$  
 
Show that, for $a < b$, 
$$ 
\int_p^q \left( x -  a \over b - x\right)^{\!\frac12} \!\d x = 
\frac{(b-a)(\pi +3{\surd3}  -6)}{12},  
$$  
where $p=  {(3a+b)/4}$ and $q={(a+b)/2}$.

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