Using the substitution `x=tan^(2)theta`, evaluate
`int_(0)^(1)sin^(-1)sqrt((x)/(1+x))\ dx` (4 marks)
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Using the substitution `x=tan^(2)theta`, evaluate
`int_(0)^(1)sin^(-1)sqrt((x)/(1+x))\ dx` (4 marks)
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`pi/2-1`
`x=tan^(2)theta`
`dx/(d theta)=2sec^2theta\ tantheta\ \ =>\ \ dx=2sec^2theta\ tantheta\ d theta`
`text{When}\ x=0, \ theta=0`
`text{When}\ x=1, \ theta=pi/4`
`sin^(-1)sqrt((x)/(1+x))` | `=sin^(-1)sqrt((tan^(2)theta)/(1+tan^(2)theta))` | |
`=sin^(-1)sqrt((tan^(2)theta)/(text{sec}^(2)theta))` | ||
`=sin^(-1)sqrt((sin^(2)theta))` | ||
`=sin^(-1)(sintheta)` | ||
`=theta` |
`int_(0)^(1)sin^(-1)sqrt((x)/(1+x))\ dx=int_(0)^(pi/4)\ theta xx 2sec^2theta\ tantheta\ d theta`
`text{Integrating by parts:}`
`u` | `=theta` | `u′` | `=1` |
`v′` | `=2tan theta\ sec^2 theta` | `v` | `=tan^(2)theta` |
`int_(0)^(pi/4)\ theta xx 2sec^2theta\ tantheta\ d theta`
`=[theta\ tan^(2)theta]_0^(pi/4)-int_0^(pi/4)tan^(2)theta\ d theta`
`=(pi/4 xx 1 -0)-int sec^(2)theta-1\ d theta`
`=pi/4-[tan theta-theta]_0^(pi/4)`
`=pi/4-[(1-pi/4)-0]`
`=pi/2-1`
Using the substitution `t=tan\ x/2`, find
`int(dx)/(1+cos x-sin x)` (3 marks)
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`-lnabs(1-tan(x/2))+C`
`text(Let)\ \ t = tan\ x/2, \ cos\ x = (1-t^2)/(1 + t^2), \ sin\ x=(2t)/(1+t^2)`
`dt = 1/2 sec^2\ x/2\ dx \ => \ d x = (2\ dt)/(sec^2\ x/2) = 2/(1 + t^2)\ dt`
`text{I}` | `= int(dx)/(1+cos x-sin x)` | |
`=int 1/(1+(1-t^2)/(1 + t^2)-(2t)/(1 + t^2)) *2/(1 + t^2)\ dt` | ||
`=int 2/(1+t^2+1-t^2-2t)\ dt` | ||
`=int 1/(1-t)\ dt` | ||
`=-ln abs(1-t)+C` | ||
`=-lnabs(1-tan(x/2))+C` |
Of the following expressions, which one need NOT contain a term involving a logarithm in its anti-derivative?
`C`
`text{Consider the denominator of}\ C:`
`x^(3)-x^(2)+x-1` | `=x^2(x-1)+(x-1)` | |
`=(x^2+1)(x-1)` |
`(x-1)/(x^(3)-x^(2)+x-1)=(x-1)/((x^2+1)(x-1))=1/(x^2+1)`
`int 1/(x^2+1)\ dx=tan^(-1)(x)+c`
`=>C`
With a suitable substitution `int_(pi/4)^(pi/3) (sec^2(x))/(sec^2(x) - 3 tan(x) + 1)\ dx` can be expressed as
`B`
`text(Let)\ \ u = tan(x)`
`(du)/(dx) = sec^2 (x) \ => \ du = sec^2(x)\ dx`
`text(When)\ ` | `x = pi/3,\ u = sqrt3` |
`x = pi/4,\ u = 1` |
`int_(pi/4)^(pi/3) (sec^2(x))/(sec^2(x) – 3 tan(x) + 1)\ dx` |
`= int_1^(sqrt3)\ 1/(u^2 + 1 – 3u + 1)\ du` |
`= int_1^(sqrt3) 1/(u^2 – 3u + 2)\ du` | |
`= int_1^(sqrt3) 1/((u – 2)(u – 1))\ du` | |
`= int_1^(sqrt3) 1/(u – 2) – 1/(u – 1)\ du` |
`=>B`
Find `int(cos theta)/(sin^5 theta) d theta` (2 marks)
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`(-1)/(4sin^4 theta ) + C`
`text(Let) \ u` | `= sin theta ` |
`(du)/(d theta)` | `=cos theta => d u = cos theta \ d theta` |
`int(cos theta)/(sin^5 theta) d theta` | `= int u^-5 d u` |
`=(-1)/(4) u ^-4 + C` | |
`=(-1)/(4sin^4 theta ) + C` |
Using the substitution `x = sin^2 theta`, or otherwise, evaluate `int_0^(1/2) sqrt(x/(1 - x))\ dx`. (3 marks)
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`pi/4 – 1/2`
`x = sin^2 theta`
`dx = 2 sin theta cos theta\ d theta`
`text(If)\ \ x = 1/2, sin theta = 1/sqrt 2, \ theta = pi/4`
`text(If)\ \ x = 0, \ theta = 0`
`int_0^(1/2) sqrt(x/(1 – x))\ dx` | `= int_0^(pi/4) sqrt ((sin^2 theta)/(1 – sin^2 theta)) xx 2 sin theta cos theta\ d theta` |
`= int_0^(pi/4) (sin theta)/(cos theta) xx 2 sin theta cos theta\ d theta` | |
`= int_0^(pi/4) 2 sin^2 theta\ d theta` | |
`= int_0^(pi/4) 1 – cos 2 theta\ d theta` | |
`= [theta – 1/2 sin 2 theta]_0^(pi/4)` | |
`= (pi/4 – 1/2 sin {:pi/2) – (0 – 0)` | |
`= pi/4 – 1/2` |
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i. `(x – 1) (5 – x)` | `= 5x – x^2 -5+ x` |
`= -(x^2 – 6x + 5)` | |
`= 4 – (x – 3)^2` | |
`= 2^2 – (x – 3)^2` |
ii. | `text(Let)\ \ x – 3` | `= 2 sin theta` |
`dx` | `= 2 cos theta\ d theta` |
`text(When)\ \ x = 1,\ \ theta = -pi/2`
`text(When)\ \ x = 5,\ \ theta = pi/2`
`:.int_1^5 sqrt (4 – (x – 3)^2)\ dx`
`=int_((-pi)/2)^(pi/2) sqrt ((4 – 4 sin^2 theta))*2 cos theta\ d theta` |
`=4 int_((-pi)/2)^(pi/2) sqrt(cos^2 theta) * cos theta\ d theta` |
`=4 int_((-pi)/2)^(pi/2) cos^2 theta\ d theta` |
`=2 int_((-pi)/2)^(pi/2) (1 + cos 2 theta)\ d theta` |
`=2[theta + (sin 2 theta)/2]_((-pi)/2)^(pi/2)` |
`=2[(pi/2 + 0) – ((-pi)/2 – 0)]` |
`=2 pi` |
Without evaluating the integrals, which one of the following integrals is greater than zero?
(A) `int_(−pi/2)^(pi/2) x/(2 + cos x)\ dx`
(B) `int_(−pi)^pi x^3 sin x\ dx`
(C) `int_(−1)^1 (e^(−x^2) − 1)\ dx`
(D) `int_(−2)^2 tan^(−1)(x^3)\ dx`
`B`
`text{Consider (A) and (D)}`
`f(x)=-f(-x)\ \ =>\ text(ODD functions where)`
`int_(−a)^a f(x)\ dx = 0`
`text{Consider (C)}`
`e^(−x^2)<1\ \ text(for all)\ x\ \ => e^(−x^2) − 1<0`
`:. text(Its graph is below the)\ x text(-axis and any integral)`
`text(will be negative)`
`text{Consider (B)}`
`text{(B)}\ text(is an even function where,)`
`x^3 sinx>=0\ \ text(for)\ \ \ -pi<=x<=pi`
`=>B`
Evaluate `int_1^sqrt 3 1/(x^2 sqrt (1 + x^2))\ dx.` (4 marks)
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`sqrt 2 – (2 sqrt 3)/3`
`text(Let)\ \ x = tan theta,\ \ \ \ dx = sec^2 theta\ d theta`
`text(When)\ \ x = 1,\ \ theta = pi/4`
`text(When)\ \ x = sqrt 3,\ \ theta = pi/3`
`int_1^sqrt 3 1/(x^2 sqrt (1 + x^2))\ dx` | `= int_(pi/4)^(pi/3) (sec^2 theta)/(tan^2 theta sqrt (1 + tan^2 theta))\ d theta` |
`= int_(pi/4)^(pi/3) (sec^2 theta)/(tan^2 theta sec theta)\ d theta` | |
`= int_(pi/4)^(pi/3) (sec theta)/(tan^2 theta)\ d theta` | |
`= int_(pi/4)^(pi/3) (cos^2 theta)/(cos theta sin^2 theta)\ d theta` | |
`= int_(pi/4)^(pi/3) (cos theta\ d theta)/(sin^2 theta)` | |
`= [-1/(sin theta)]_(pi/4)^(pi/3)` | |
`= (-2/sqrt 3 + sqrt 2)` | |
`= sqrt 2 – (2 sqrt 3)/3` |
Evaluate `int_0^(pi/4) tan\ x\ dx`. (3 marks)
`1/2 ln 2 \ \ text(or)\ \ ln\ sqrt2`
`int_0^(pi/4) tan\ x\ dx` | `=int_0^(pi/4) (sin\ x)/(cos\ x)\ dx` |
`=[-ln\ cos\ x]_0^(pi/4)` | |
`=[-ln\ cos\ pi/4 – (-ln cos 0)]` | |
`=-ln\ 1/sqrt2 + ln\ 1` | |
`=ln sqrt2` | |
`=1/2 ln 2` |
Let `I = int_1^3 (cos^2(pi/8 x))/(x(4 - x))\ dx.`
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i. `text(Let)\ \ u = 4 – x,\ \ du = -dx,\ \ x = 4 – u`
`text(When)\ \ x = 1,\ \ u = 3`
`text(When)\ \ x = 3,\ \ u = 1`
`I` | `=int_1^3 (cos^2(pi/8 x))/(x(4 – x))\ dx` |
`=int_3^1 (cos^2 (pi/8(4 – u)))/((4 – u)u) (-du)` | |
`=-int_3^1 (cos^2(pi/2 – pi/8 u))/((4 – u)u)\ du` | |
`=int_1^3 (sin^2(pi/8 u))/(u (4 – u))\ du` |
ii. `text{S}text{ince}\ \ int_1^3 (cos^2(pi/8 x))/(x(4 – x))\ dx = int_1^3 (sin^2(pi/8 x))/(x(4 – x))\ dx`
`text(We can add the integrals such that)`
`2I` | `=int_1^3 (cos^2(pi/8 x))/(x(4 – x)) dx + int_1^3 (sin^2(pi/8 x))/(x(4 – x)) dx` |
`=int_1^3 1/(x(4 – x)) dx` |
`text(Using partial fractions:)`
`1/(x(4-x))` | `=A/x+B/(4-x)` |
`1` | `=A(4-x)+Bx` |
`text(When)\ \ x=0,\ \ A=1/4`
`text(When)\ \ x=0,\ \ B=1/4`
`2I` | `=1/4 int_1^3 (1/x + 1/(4 – x))\ dx` |
`=1/4 [log_e x – log_e (4 – x)]_1^3` | |
`=1/4 [log_e 3 – log_e 1 – (log_e 1 – log_e 3)]` | |
`=1/2 log_e 3` | |
`:.I` | `=1/4 log_e 3` |