19for the first one, differentiate w.r.t (a) and then (b) and then integrate !
Q1. \int_{0}^{\pi /2}{ln(a^2cos^2\theta +b^2sin^2\theta )d\theta }
Q2. \int_{0}^{\infty }{\frac{ln(1+a^2x^2)}{1+b^2x^2}dx }
Q3. If lxl < 1 then find the sum of the series \frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}+...
Q4.\int_{0}^{\pi /4}{\frac{x^2(sin2x-cos2x)}{(1+sin2x)cos^2x}}
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14 Answers
19
106i couldnt get u .. U want me to differentiate that wrt a ? even when another variable θ is der? do i leave it as a constant??
19standard property :
if a function f(x,α) be continuous for a≤x≤b and c≤α≤d , then for any α ε [c,d], if I(α) =a∫b f(x,α) dx ,,,then d/dα { I(α)}= a∫b δ/δα {f(x,α)} dx
341This should help: http://targetiit.com/iit-jee-forum/posts/prove-the-following-11283.html
21sorry bud. but i dun hav the patience to type such a big soln here......tell me wer u got stuck in it...
1357For third
S=given expression
∫S=ln(1+x)+ln(1+x2)....+C
∫S=ln[(1+x)(1+x2)(1+x4)......]+C
∫S=ln(1+x+x2+x3........)+C
∫S=ln[1/(1-x)]+C=-ln(1-x)+C
differentiating
S=1/(1-x)
3413rd one(w/o calculus): If the given sum is S what is S - 1/1-x?
1q-2)
\frac{\partial I(a)}{\partial a}=\int_{0}^{\infty}\frac{2ax^2\mathrm{dx}}{(1+a^2x^2)(1+b^2x^2)}
I'(a)=\frac{\pi}{ab+b^2}\\ I(a)=\frac{\pi}{b}\int \frac{\mathrm{da}}{a+b}\\ I(a)=\frac{\pi}{b}\left ( \ln(a+b) +c \right )\\ I(0)=0\\ c=-\ln b\\ I(a)=\frac{\pi}{b}(\ln(\frac{a+b}{b}))
i know ppl will call me a lunatic idiot, but still i had wanted to solve this question wen it was posted...and asish please tell if the answer is right ??
106how m i supposed to know da answr now ?? (it was asked in Dec 09) :O .. further im not into JEE syll anymore(and hv frgtten hw 2 integrate)
but your steps seem correct. .. so the answer also must be so
- Asish Mahapatra looking at the english I was using 2 years ago makes me cringe now. And sorry for grave-digging a 2yr old thread :(Upvote·0· Reply ·2013-05-26 05:38:03
1thats correct bindaas. .. substitutin a=1,b=1 we get a familiar definite integral
\int_{0}^{\infty}\left ( \frac{ln(1+x^2)}{1+x^2} \right )= \pi log2
hence i guess answer is write