12 Show that \(\int _ { 0 } ^ { \frac { 1 } { \sqrt { 3 } } } \frac { 4 } { 1 - x ^ { 4 } } d x = \ln ( a + \sqrt { b } ) + \frac { \pi } { c }\) where \(a , b\) and \(c\) are integers to be determined.
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Question 12:
= + +
1−x4 1−x 1+x 1+x2
⇒ A(1+x)(1+x2)+B(1−x)(1+x2)+(Cx+D)(1−x2)=4
x=1:4A=4⇒ A=1
x=−1:4B =4⇒ B =1
x=0:A+B+D =4⇒ D =2
x3:A−B−C =0⇒C =0
1
3 1 1 2
I = ∫ + + dx
1−x 1+x 1+x2
0
1
1+x 3
= ln +2tan−1x
1−x
0
3+1 1 3+1 π
=ln +2tan−1 (−0 ) =ln +
3−1 3 3−1 3
( ) π
=ln 2+ 3 + i.e. a =2, b=3, c=3
A1
A1
M1
M1
1.1
1.1
1.1
1.1
Answer Marks
1.1 Proper split to produce integrable integrand
For one of the terms /constants
For all terms/constants
1
Or of these
4
See below for other possibilities
Correctly integration of their integrand
without simplification – ignore limits
Substitution – ignore – 0
Values must be stated
[6]
= + +
Answer Marks
Guidance
1−x4 1−x 1+x 1+x2 M0
But consider last 3 marks for correct
integration
4 A D
= + with A = D = 2 by inspection
1−x4 1−x2 1+x2
2 1 1
Followed by + + in integration section
Answer Marks
1+x2 1−x 1+x M1
A1
Answer Marks
A1 Look to see the second split further on in
question
4 A D
= +
1−x4 1−x2 1+x2
1 1
3 4 3 2 2
⇒ ∫ dx= ∫ + dx
1−x4 1−x2 1+x2
0 0
1 1 1
= 2tanh−1x+2tan−1x 3 = 2tanh−1 +2tan−1
0 3 3
1
1+
3 π
=ln +2 −0
1 6
1−
3
3+1 π 1+2 3+3 π 4+2 3 π
=ln + =ln + =ln +
3−1 3 2 3 2 3
( ) π
=ln 2+ 3 +
A1
Answer Marks
A1 Look for the integration. If tanh-1 is used
then give full marks here.
PMT
OCR (Oxford Cambridge and RSA Examinations)
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Shaftesbury Road
Cambridge
CB2 8EA
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Education and Learning
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Facsimile: 01223 552627
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Copy
Question 12:
12 | 4 A B Cx+D
= + +
1−x4 1−x 1+x 1+x2
⇒ A(1+x)(1+x2)+B(1−x)(1+x2)+(Cx+D)(1−x2)=4
x=1:4A=4⇒ A=1
x=−1:4B =4⇒ B =1
x=0:A+B+D =4⇒ D =2
x3:A−B−C =0⇒C =0
1
3 1 1 2
I = ∫ + + dx
1−x 1+x 1+x2
0
1
1+x 3
= ln +2tan−1x
1−x
0
3+1 1 3+1 π
=ln +2tan−1 (−0 ) =ln +
3−1 3 3−1 3
( ) π
=ln 2+ 3 + i.e. a =2, b=3, c=3
3 | M1
A1
A1
M1
M1
A1 | 3.1a
1.1
1.1
1.1
1.1
1.1 | Proper split to produce integrable integrand
For one of the terms /constants
For all terms/constants
1
Or of these
4
See below for other possibilities
Correctly integration of their integrand
without simplification – ignore limits
Substitution – ignore – 0
Values must be stated
[6]
12 | 4 A B D
= + +
1−x4 1−x 1+x 1+x2 | M0 | But consider last 3 marks for correct
integration
4 A D
= + with A = D = 2 by inspection
1−x4 1−x2 1+x2
2 1 1
Followed by + + in integration section
1+x2 1−x 1+x | M1
A1
A1 | Look to see the second split further on in
question
4 A D
= +
1−x4 1−x2 1+x2
1 1
3 4 3 2 2
⇒ ∫ dx= ∫ + dx
1−x4 1−x2 1+x2
0 0
1 1 1
= 2tanh−1x+2tan−1x 3 = 2tanh−1 +2tan−1
0 3 3
1
1+
3 π
=ln +2 −0
1 6
1−
3
3+1 π 1+2 3+3 π 4+2 3 π
=ln + =ln + =ln +
3−1 3 2 3 2 3
( ) π
=ln 2+ 3 +
3 | M1
A1
A1 | Look for the integration. If tanh-1 is used
then give full marks here.
PMT
OCR (Oxford Cambridge and RSA Examinations)
The Triangle Building
Shaftesbury Road
Cambridge
CB2 8EA
OCR Customer Contact Centre
Education and Learning
Telephone: 01223 553998
Facsimile: 01223 552627
Email: general.qualifications@ocr.org.uk
www.ocr.org.uk
For staff training purposes and as part of our quality assurance programme your call may be
recorded or monitored
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12 Show that $\int _ { 0 } ^ { \frac { 1 } { \sqrt { 3 } } } \frac { 4 } { 1 - x ^ { 4 } } d x = \ln ( a + \sqrt { b } ) + \frac { \pi } { c }$ where $a , b$ and $c$ are integers to be determined.
\hfill \mbox{\textit{OCR Further Pure Core 1 2020 Q12 [6]}}