07-09-09 Integration .. Use the last trick we learnt!

\int \frac{x^2}{(x^2-4)\sin x+4x \cos x}dx

5 Answers

same trick
(x^2-4)\sin x+4x\cos x=(x^2+4).cos(x-\tan ^{-1} \frac{x^2-4}{4x})

Putting x-\tan ^{-1} \frac{x^2-4}{4x}=t

=> (1-4\frac {x^2+4}{(x^2-4)^2})dx=dt

acha bhaiya...how do we knw wen to use dis trick??

Practice :P

btw .. good work eureka..

can someone give teh final answer?

Solution contd..

[ 64 x² / (x² + 4)] dx = dt

I = 1/64 integ(sec t dt)

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