If the above integral equals , where and are positive integers, find .
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
K = ∫ 1 2 csc ( 2 x ) [ l n ( s i n 2 ( x ) ) − l n ( c o s 2 ( x ) ) ] 2 d x
Using properties of logarithm ,
l n ( s i n 2 ( x ) ) − l n ( c o s 2 ( x ) ) = ln tan 2 x
Thus our integrand becomes
K = ∫ 1 2 csc ( 2 x ) [ ln tan 2 x ] 2 d x
Now let t = l n tan 2 x
d x d t = 4 csc 2 x
Therefor K = ∫ 3 t 2 d t
Which on evaluating and substituting back t = ln tan 2 x gives K = ln 3 tan 2 x
On putting limits and evaluating, final answer becomes ln 3 3