11Strider koi question pooch
usse samajhenge[1]
This thread features my doubts in integration problems.....others can feel free to post their own!!!
OK HERE GOES:
I came across these questions in Brilliant Tutorials Elite Course ......
(I)
Both of the questions below have one or more options correct, but funnily enough i was able to pick out only one correct option in each question....
Pls give all possible correct options for both questions:
1)\int \frac{2x.dx}{1+x^4}
(a) tan-1(x2)+c (b)1/2 tan-1(x2)+c
(c)log(1+x4)+c (d) tan-1(1/x2)+c
2)\int \frac{dx}{x+\sqrt{x}}
(a)log(1+√x)+c (b) 1/2 log(1+√x)+c
(c)2log(1+√x)+c (d)-1/2 log(1+√x) +c
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32 Answers
16)
(5-4x-x2)5/2 = (5+x)5/2(1-x)5/2
now write numerator as : 1/6[(5+x) + (1-x)]
then the integration bcmes:
1/6 [ ∫(5+x)/(5+x)5/2(1-x)5/2 dx + ∫(1-x)/(5+x)5/2(1-x)5/2 dx. ]
=1/6[ ∫dx/(5+x)3/2(1-x)5/2] + ∫dx/(5+x)5/2(1-x)3/2 ]
now apply the trick of maverick...
remeber that is a standard trick ...
118) put x =tanθ
and dx =sec2θdθ
and put y =sin-1p
and dy =1/√1-p2dp
naw change the limits and the question is done[1]
11PLEASE COMPARE THE QUESTIONS WITH JEE LEVEL OF TOUGHNESS AND THEN SOLVE:
#5)
(Subjective again)
\int \frac{(x-1)\; dx}{(x+1)\sqrt{x^3+x^2+x}}
#6)
\int \frac{dx}{\left(5-4x-x^2 \right)^{\frac{5}{2}}}
#7)
\int \frac{dx}{(x-1)^\frac{3}{4}(x+2)^\frac{5}{4}}}
#8)
Prove that
\int_{0}^{1}{\frac{tan^{-1}x\; dx}{x}}=\int_{0}^{\frac{\pi }{2}}{\frac{y\; dy}{siny}}
11[1]
Thanx da
So everytime, i get a question like this...... i'll remember u da. [1]
11YUP ARAGON I THINK U GOT IT NOW
FOR UR INFORMATION U SOLVED A QUES OF IIT 1997 [1]
11that'll become......
\left[\sqrt{1+t^2} \right]^{2}_{0}
Hence....(√5-1)??
11SOLVE THE FULL QUESTION
UWIONT GET ANYTHING FOR HALF SOLUTION[2]
DO IT THEN WE WILL CHECK[1]
11putting r/n---->t, dr/n----dt
It became
\int_{0}^{2}{\frac{t\; dt}{\sqrt{1+t^2}}}
Correct till now??
11\lim_{n->infinity} 1/n\sum_{r=1}^{2n}{}r/\sqrt{n^{2}+r^{2}}
THIS ONE IS FOR ANIRUDH ONLY!!!!!!!!
11yes i may say
in most of the other cases u have to struggle hard to find it
but it of premeir importance
chal i give u a ques 4 practice
11What should we look to do in such questions??
Always substitute r/n--->t??
11yaar dekh iss mein
put r/n------>t
then 1/n-------->dt
this could be represented in the form of integral
lower limit and upper limit can be found out in the following method
For lower limit
sigma ke nicche waali term check kar
aggar woh constant hai to u put lower limit as 0
For upper limit
check the coefficient of n in the upper power of sigma
the coefficient must be written in the upper limit
so
noe the ques becomes
∫01x99dx
and hence 1/100
[1]
11Pls post complete solution with easy-to-understand reasoning
Exclude terminologies as much as possible
11OK then....
\lim_{n\rightarrow \infty}\frac{1}{n^{100}}\sum_{r=1}^{n}{r^{99}}
11(II)
Subjective type
(1)\int sin^3x\; cos(\frac{x}{2})dx
(2)\int sin^{-1}\left(\sqrt{\frac{x}{a+x}} \right)dx
(3)\int \frac{dx}{sinx(3+2cosx)}
(4)\int \frac{sinx\; dx}{sin3x}
11Pls give me the algorithm for converting summation to a definite integral...
Also pls illustrate with an example!
1@manipal.... afetr where i left it .. in 2nd questions...
its jus one step...
becoz tanθsec2θdθ taken as second function... a child can integrate..
anyways... u have given another method... dat way its nice :)
1arey no no... all those partial fractions will become by parts !! [3] sorry..
1subj. 3)
∫dx/sinx(3+2cosx)
= ∫sinxdx/sin2x(3+2cosx)
= ∫sinxdx/(1-cos2x)(3+2cosx)
now put cosx=t...
11Q2
I THINK IT WOULD BE BETTER TO PUT
x+a=t2
dx=2tdt
upper niche waala t kat jaaega
so the ques becomes
∫sin-1√t2-adt
this could be found out by PARTS (EDITED)[1]
14) ∫sinx/sin3x = ∫sinx/sinx[3-sin2x] = ∫1/[2+cos2x]
=∫sec2xdx/[2sec2x+1]
=∫sec2xdx/[3+2tan2x]
tanx =t le lena ......
then it becomes simple quadratic..
1subjective:
2) sin-1√x/(x+a)
take x=atan2θ => dx = 2atanθsec2θdθ
∫sin-1√x/(x+a)
=∫sin-1(sinθ) .2atanθsec2θdθ
= 2a∫θtanθsec2θdθ
now do partial fractions taking θ as first function and tanθsec2θ as second function.
final ans : a[((x/a) +1)tan-1√x/a - √x/a ]
1subjctive tpe
3rd
tan(x/2)=t
proceed aajayega
ni aaya toh kck knck
1for 2nd q put x=a tan 2θ then u ll get 2aθtanθsec2θthen use by parts taking θ as 1st fn n rest as 2nd i think it ll do bt still better approaches invited
1for 1 subjective problem put sinx=2sinx/2 cosx/2 n then put cos x/2 as t