If #u_n = int (sin nx)/sinx dx, >= 2#, prove that #u_n = (2sin(n-1)x)/(n-1)+u_(n-2)# Hence evaluate: #int_0^(pi/2) (sin5x)/ sinx dx#?
4 Answers
Explanation:
Calling
Here we used
with
and now
so
Proof is given in the Explanation.
Explanation:
In what follows,
Hence, the Proof.
Using
And, finally, without
Enjoy Maths.!
For the
Explanation:
Here is a Second Method to prove the Result for
Now,
Knowing that,
Rest of the Soln. is the same as in the First Method.
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Explanation:
Without the use of the given Recurrence Relation,
A Result :
We have,
Using the above Result,
Adding
Since,
we get,
Enjoy Maths.!