mixture: March 2013

Monday, March 11, 2013

integral using manipulation of numerator in terms of denominator

integral of 1/[sin²x cos²x]

since sin²x + cos²x =1
rewrite the numerator as sin²x + cos²x
divide each term with the denominator
simplify and then integrate

study integration formulae

free guide on integration or integral calculus for cbse isc kerala higher secondary class 12 ncert scert students

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integral using substitution

integral of [√(tanx)] / [sinx cosx]

first introduce [√(tanx)] in the numerator and denominator

then simplify and change the tanx in the numerator in terms of sinx and cosx
simplify in terms of secx and tanx

then use substitution method

study integration formulae

Saturday, March 9, 2013

integral using trigonometric manipulation

integral of 1 / [cos(x-a)cos(x-b)]

we will try to write the numerator in terms of the denominator

so introduce sin(a-b) in the numerator and denominator

note that (x-b)-(x-a) =a-b

so write sin(a-b) as sin[ (x-b)-(x-a) ] and expand using sin[A-B] formula

divide the numerator term by term with the denominator

reduce to terms involving sec and then integrate

study integration formulae a

integral 5

integral of 1 / [x - √x ]

we have to use substitution method

take √x common in the denominator

put √x =t
study differentiation formulae ands integration formulae
then take the differential

(1/[2√x] ) dx = dt

rearrange the integral and rewrite it in terms of t in a single step
and then integrate

note that log x stands for natural logarithm ln x in this question

integral 4

integral of (log x )² / x
note that log x stands for natural logarithm ln x in this question

since there is no quotient rule in integration and since this is not a standard form you have to try to use substitution method to get to a standard form

so you put log x =t then take the differential

so (1/x) dx = dt

after that arrange the integral so that you can rewrite the integral in terms of t in a single step

then you get a standard integral which can be integrated.
study integration formulae

integral 3

integral of sec²x/cosec²x

since there is no quotient rule in integration, change everything in terms of sinx and cosx simplify to get tan²x
since there is no direct integral for tan²x change to sec²x-1 using trigonometry formulae and then integrate term by term.
study integration formulae
and

trigonometry identities formulae etc

integral 2

integral of [ x^3 - x^2 + x -1] / (x-1)

factorise the numerator

cancel off the common factor

use the integral of x^n to integrate term by term
go through the integration formulae

Friday, March 8, 2013

integral1

integral of (x^3 +3x +4) /sqrt(x)

since quotient rule is not used in integration you divide each term in the numerator with the denominator and
reduce every thing to x^n form and then integrate term by term
go through the integration formulae

Thursday, March 7, 2013

solving a equation involving exponentials

solve sinh (x)=0.5
(eˣ - e⁻ˣ)/2 =0.5 hyperbolic functions
so that eˣ - e⁻ˣ =1

now solve eˣ - e⁻ˣ = 1

multiply with eˣ

make it into a quadratic in eˣ

solve using quadratic fomula to find the value of eˣ

take natural logarithm to find x

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index of math problems

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Wednesday, March 6, 2013

proving a trigonometric identity

proving a trigonometric identity

change everything in terms of sinx and cosx
simplify, then use trigonometric identities

trigonometry identities formulae etc

integral using trigonometric substitution

integral of (x^3)/ [sqrt[(x^2)+4] ]
use the substitution x = 2tanθ
dx = 2 ((secθ)^2) dθ
simplify and then use another substituion
secθ = t
secθ tanθ dθ = dt
simplify and integrate then resubstitute
change secθ in terms of tanθ
then again resubstitute
remembering that x =2 tanθ or tanθ = (x/2)

integration formulae

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index of math problems

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There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work

Tuesday, March 5, 2013

solving a second order differential equation using reduction of order method if one of the solutions is given

solve y"-y = 0 if y=coshx is one of the solutions
using the formula for reduction of order
substitute p=0 since there is no term in y' and
use y1 = cosh x the given solution simplify and then integrate
integration formulae

Monday, March 4, 2013

problem on scalar triple product stp of vectors

show that [axb bxc cxa] = [a b c]^2
where a,b,c are vectors

use the formula for crossproduct of four vectors

then use the property that if a vector is repeated in an stp, then the value of the stp is zero

then use the property that interchange of vectors in an stp changes the sign of the stp.

some questions in mathematics for the tamilnadu higher secondary 12th exam, isc , puc pre degree etc.
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index of math problems

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Friday, March 1, 2013

to find the equation of the tangents that can be drawn from a given point to a given hyperbola in standard form

find the equation of the tangents from (1,2) to the hyperbola 2(x^2) -3(y^2) =6

divide by the term on the rhs to reduce the given equation to standard form
extract the value of a and b
use the equation of the tangent to the hyperbola in standard form and use the values of a and b
use the condition that the tangent passes through the given point to form an equation in m and solve for m
and resubstitute

solving a fourth degree equation with real coefficients if you are given one of the imaginary roots

solving a fourth degree equation with real coefficients if you are given one of the imaginary roots

solve x^4 - 4 x^3 + 11 x^2 -14x +10 =0 given that 1+2i is one of the roots.

since the coefficients of the given equation are real, the complex roots will occur in conjugate pairs

so 1-2i is another root.

using this you can find a quadratic factor of the given fourth degree expression.

use long division or comparision of coeffecients to extract the other quadratic factor and solve it

checking consistency of equations using determinants

Solve using determinants (Cramers rule)

2x + 3y = 8

4x + 6y = 16

n = 2 unknowns
r =1

n-r = 2-1 = 1

therefore one unknown can be chosen arbitrarily

put y = k (arbitrary ) in the first equation 2x +3y =8

2x + 3k = 8

solving x = (8 - 3k) / 2

so solution is

x = (8 - 3k) / 2

y = k

where k is any number.

some questions in mathematics for the tamilnadu higher secondary 12th exam, isc , puc pre degree etc.
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please leave your comments below

------------------------------------------------------------

index of math problems

disclaimer:

There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work

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