help...

[hassam]

 

[destined007]

None of these. Its answer is pi/4. Its a very complex integration.

 

[destined007]

where did you find this question anyway?

 

[hassam]

its an entrance test probably indian....PROBLEM....I cnt undrstnd how u got anser.

 

[hassam]

chek this http://www.wolframalpha.com/input/?i=integrate+%28x^2-1%29%2F%28x^3sqrt%28x^2-1%29%29

 

[destined007]

yeah. You won't get this kinda question in A level or any university's aptitude test.

 

[destined007]

yeah. You won't get this kinda question in A level or any university's aptitude test.

 

[hassam]

cud u solve....i solved it now...

 

[hassam]

well there is one more problem shud i post....?

 

[destined007]

yeah sure.

 

[hassam]

here u go

 

[destined007]

Its c.

 

[usman]

@ Hassam:- U can differentiate each of the expressions in the given options w.r.t. x, keeping in mind that d/dx [ tan^-1(x) ] = 1/( 1+x^2 ) and then check which of the resullts of the differentiated options simplies to the given integrand. U'll find that it's C

 

[usman]

Salam. Hassam yar I tried to solve the above integral without taking help from the given options. Here’s my solution (Alternative to but a little brief than the Wolfram Alpha’s one):-

Int < [ cos^3(x) + cos^5(x) ] / [ sin^2(x) + sin^4(x) ] dx >
= Int < [ cos^3(x) + cos^5(x) ] / sin^2(x)*[ 1 + sin^2(x) ] dx >

Let’s try to resolve the integrand into partial fractions in order to simplify the integration process.
Now, we observe that the numerator (degree 5) of the integrand (which is a fraction) has a degree one higher than the denominator (degree 4), considering “sin(x)” as “X” while “cos(x)” as “sqrt(1 – X^2),” or vice versa. So, the integrand can be resolved to a form such as:-

A / sin^2(x) + B / [ 1 + sin^2(x) ] + C = Integrand = [ cos^3(x) + cos^5(x) ] / sin^2(x)*[ 1 + sin^2(x) ]
where A, B and C are functions of x (Most probably trigonometric ones).

We can multiply the above identity throughout by “sin^2(x) [ 1 + sin^2(x) ].” Doing so leads to:-

A*[ 1 + sin^2(x) ] + B*sin^2(x) + C*sin^2(x)*[ 1 + sin^2(x) ] = cos^3(x) + cos^5(x)

Now, there are two possible ways (of which I know about) for determining A, B and C. The first involves trying to express every term in the above identity in terms of sin(x) OR secondly in terms of cos(x).

Let’s use the 1st way firstly ---->

A*[ 1 + sin^2(x) ] + B*sin^2(x) + C*sin^2(x)*[ 1 + sin^2(x) ] = cos^3(x) + cos^5(x)
= cos(x)*[ cos^2(x) + cos^4(x) ]
= cos(x)*[ 1 – sin^2(x) + { 1 – sin^2(x) }^2 ]
= cos(x)*[ 2 – 3sin^2(x) + sin^4(x) ]

Expanding both the L.H.S. and R.H.S., and taking “sin^2(x)” common on L.H.S. leads to---->

A + (A + B + C)*sin^2(x) + C*sin^4(x) = 2cos(x) – 3cos(x)*sin^2(x) + cos(x)*sin^4(x)
Equating the two left-most terms on each side and the coefficients of sin^2(x) and sin^4(x), followed by solving the resulting equations simultaneously, we get:-
A = 2cos(x); B = - 6cos(x); C = cos(x)

Now, let’s use the 2nd way (in order to check the consistency of results):-

A*[ 1 + sin^2(x) ] + B*sin^2(x) + C*sin^2(x)*[ 1 + sin^2(x) ] = cos^3(x) + cos^5(x)

A*[ 1 + 1 - cos^2(x) ] + B*[ 1 - cos^2(x) + C*[ 1 - cos^2(x) ]*[ 1 + 1 - cos^2(x) ] = cos^3(x) + cos^5(x)
Simplifying and expanding the above expression, followed by taking “cos^2(x)” common on L.H.S. and writing “cos^3(x) + cos^5(x)” on L.H.S. as “cos(x)*cos^2(x) + cos(x)*cos^4(x),” we get:-

(2A + B + 2C) + (- A – 3C)*cos^2(x) + C*cos^4(x) = 0 + cos(x)*cos^2(x) + cos(x)*cos^4(x)
Equating the two left-most terms on each side and the coefficients of cos^2(x) and cos^4(x), followed by solving the resulting equations simultaneously, we get:-
A = 2cos(x); B = - 6cos(x); C = cos(x)

Bravo! @ the results for A, B & C do match….! We are now half-way through. We have successfully converted the integrand from a complicated form:-
Int < [ cos^3(x) + cos^5(x) ] / [ sin^2(x) + sin^4(x) ] dx >

To a much simpler, more workable form:-

Int < 2cos(x) / sin^2(x) – 6cos(x) / [ 1 + sin^2(x) ] + cos(x) dx >

= 2*Int < cos(x) / sin^2(x) dx > – 6*Int < cos(x) / [ 1 + sin^2(x) ] dx > + Int < cos(x) dx >

Let’s substitute “sin(x)” by “u,” in the first and the second integrands in the above expression. Then du/dx = cos(x) = sqrt[ 1 – sin^2(x) ] = sqrt[ 1 – u^2 ] -----> dx = du / sqrt[ 1 – u^2 ].
Therefore, also substituting “cos(x)” by “sqrt[ 1 – u^2 ]” and “dx” by “du / sqrt[ 1 – u^2 ].”

2*Int < sqrt[ 1 – u^2 ] / u^2 * du / sqrt[ 1 – u^2 ] > – 6*Int < sqrt[ 1 – u^2 ] / [ 1 + u^2 ] * du / sqrt[ 1 – u^2 ] > + Int < cos(x) dx >
= 2*Int < du / u^2 > – 6*Int < du / [ 1 + u^2 ] > + Int < cos(x) dx >
= - 2/u – 6arctan(u) + sin(x) + constant
where arctan(u) = tan^-1(u)

Now, finally let’s back-substitute “u” by “sin(x)” ----->

= - 2 / sin(x) – 6arctan[ sin(x) ] + sin(x) + constant (i.e. part C)
= The value of the original Integral

Bingo……….!!!


Hope that helps.....

 

[princesszahra]

well there is one more problem shud i post....?

go ahead!! n ppl let me answer that one!

 

[destined007]

Actually, he already posted the 'another' problem. Hehe!

 

[princesszahra]

Actually, he already posted the 'another' problem. Hehe!

where exactly?!??!

 

[destined007]

here u go

Here it is. Above few posts.