IIT JEE , AIEEE Forum - Mathematics , Integral Calculus - Integrals Challenge : rates for best answer

hsbhatt

I think the method would be something like this:

1/1+sinx+sin²x dx

We have 1+sinx+sin²x = (1-

sinx) (1-

²sinx)

Hence 1/1+sinx+sin²x = 1/(1-

sinx) (1-

²sinx) = A/(1-

sinx) + B/(1-

²sinx)

Comparing LHS and RHS we get
A+B = 1 and
A

²+B

= 0

Eqn 2 gives on multiplying by

². A

+B = 0 or A

+1 = A giving
A = 1/1-

and B = -

/1-

Now the integral can be evaluated using the integral

dx/a+bsinx which I've unfortunately forgotten how to do.I leave it for the students to finish.

PS: I should have said: "The rest is left as an exercise for the student".

konichiwa2x

brilliant sir! I must say, that was a real peice of beauty. far better than the method I had in mind!

Shall I post my solution or are there any other takers?

elastiboysai

Excellent method sir!!!!

sboosy

Bhattji at ur very best superb

studyid

hello friends...

i did not understand the method given by hs bhatt...

can neone explain it to me.....

rates assured..

konichiwa2x

which part wasnt clear? the latter portion and the complex roots part?

nudge me abt ur doubt.

aankurverma

well 1st method is

numerator me

sinx add subtarct karke solve kar lo

n 2nd wo partial fractions waala jisme

Anumr + Bden + C waala

3rd

cos^x se divide karo den

den fir se 1st method waali situation aa jayegi add subtarct tan^2x

got it try kar lo yaar ho jayega

ni aaye toh batana wld think of ny more

konichiwa2x

no aankurverma, those wont work. please obtain the final answer before posting your methods.

punnima

ramkumar_november

guys , i found out another method....... check whether its correct........

I =

dx
-----------------------------------
1 + sinx + sin²x

I =

      sec²x   dx
         -----------------------------------------------
           sec²x   + tan²x   + secxtanx

I =

   secx.secx.dx
       -----------------------------------------------------
         (secx + tanx)²   - secxtanx

now put secx + tanx = t ..................(i)

           secx(secx+tanx)dx=dt

              secxdx=dt/t

also    secx - tanx =1/t . ......................(ii)

adding (i) and(ii)

secx = (t² + 1)/2t    and tanx = (t² - 1)/2t

so secxtanx = (t⁴ - 1)/4t²

I =

   (t²+1)dt
      -----------------------------------------
        t*2t*{ t²   - (t⁴ - 1)/4t² }

I =

2(t²+1)dt
-----------------------
3t⁴ + 1

after which i think you can solve.......................

konichiwa2x

nice one ramkumar_novemeber! In fact, the first time I tried this problem I tried to solve through your line of thinking. I tried so many substitutions, but didnt think of secx - tanx = 1/t. Simple but elegant!

konichiwa2x

Here is a more detailed form of hsbhatt sir's solution. Do nudge me if you are still finding difficulties.

is a quadratic in Sin{x}.

Solving we get,

.

The expression on the right represents the cube roots of unity represented by 1,

and

.

Hence,

Now we employ method of partial fractions to obtain,

Compare the coefficients to get,

and

.

Solve, u'll get

and

.

So the integral now becomes,

Substitute

. So that

and

.

The above two integrals are of standard form and can be solved quite easily!

konichiwa2x

Thanks to all who replied. Here is my solution:

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

Then

Let

then

Now let

and

Then:

Now just substitute back and you're done!

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