Calculust - 3
∫ π / 4 π / 3 ( sin θ ) 2 0 0 9 ( cos θ ) 2 0 0 9 ( sin 3 θ − cos 2 θ − cos 3 θ ) ( sin θ + cos θ + cos 2 θ ) 2 0 0 7 d θ
If the above integral is of the form d ( a + b ) n − ( 1 + c ) n ,
where a , b , c and d are positive integers, find a + b + c + d .
The answer is 2021.
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Beautiful problem!. seemingly difficult but on solving it turns out to be easy!.
The trick is to break the denomiator into powers viz 2007 and 2 .
Take the term (sinxcosx)^2007 into the second bracket of numerator and power 2 term into first bracket of numerator.
In the second bracket you will get (secx+cosecx+cotx)^2007 and in first bracket you will get (secxtanx-cosecxcotx-cosec^2x)
Which is nothing but the derivative of second bracket.
We make the substitution secx+cosecx+cotx = t
The integral is simply t^2007dt which is now a child's play!
Keep posting