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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  







 

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


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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 







 

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


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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






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


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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





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


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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





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


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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 







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


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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 










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


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Thursday, March 7, 2013

solving a equation involving exponentials

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

 now solve eË£ - e⁻Ë£ = 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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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


 










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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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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


 
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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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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


 
 

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






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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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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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