Calculus - very tough integral ---dare to try this one -----

6

Kalyan IIT-K Beware I'm coming ·Jan 21 '10 at 8:04

arrey aishwarya u r gettin an ans atleast m nd stuck up hw to break it......

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eragon24 _Retired ·Mar 6 '10 at 1:10

one way is to put $a s i n x = b c o s x . t a n θ$

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1

" ____________ ·Mar 6 '10 at 0:56

TAPAS YAAR MUJHE YE

" WOLFRAM MATHEMATICA " WAALA ANSWER NAHI CHAHIYE !

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Tapas Gandhi ·Mar 5 '10 at 22:50

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" ____________ ·Mar 5 '10 at 21:16

any diiferent approach to this question !!????

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1

akari ·Jan 22 '10 at 3:46

yes bhaiya
i just wanted to write
some constant k such that
kt=z
actually i didnt mean the previous constant k (soryy for dat )

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62

Lokesh Verma ·Jan 22 '10 at 3:43

i think you should use the substitution √k t = z

Otherwise it seems fine.. [1]

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akari ·Jan 21 '10 at 14:35

not quite sure
done it roughly
$\int \frac{(1+\tan^2x)\sec^2dx}{\left(a^2\tan^2x+b^2 \right)^2} \\ \\ put \ tanx =t \\ \int \frac{\left( 1+t^2\right).dt}{\left( kt^2 +1\right)^2 }\\ \int \frac{dt}{\left( kt^2 +1\right)^2} +\int \frac{t^2.dt}{\left( kt^2 +1\right)^2}\\ put \ k_2t=z \\ such \ that \ the \ integral \ becomes \\ \int k_1\frac{dz}{\left( z^2 +1\right)^2} + k_2\int \frac{z^2.dz}{\left( z^2 +1\right)^2}\\ now \ put \ z=\tan\alpha \\ k_1\int \cos^2\alpha + k_2\int \sin^2 \alpha$

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Aishwarya Maurya ·Jan 21 '10 at 8:05

All right the correct answer is complicated as well.
http://www.wolframalpha.com/input/?i=integrate+1%2F%28%28a^2+%28sin+x%29^2+%2B+b^2%28cosx%29^2%29%29^2+dx+
but yea ..it is integrable