If
∫
π
/
6
π
/
3
sec
x
+
csc
x
d
x
can be represented in the form
2
a
−
b
+
2
c
ln
(
d
+
e
−
f
−
g
)
then find
a
+
b
+
c
+
d
+
e
+
f
+
g
Detials and assumptions
1) a , b , c , d , e , f , g are integers all less than 1 0 . They need not to be distinct
2) a , c , d , e , f are not having any factor which is a perfect square.
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"This is how it's done"( Courtesy : Counter Strike condition zero )
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:D You're an avid Gamer, it seems!
Try this: . I tweaked this problem inspired by a problem similar to yours.
Same method as mine!!!! At end I further simplified the last term as
− 2 1 c o s x + 2 1 s i n x − 2 2 1 l n ∣ t a n ( 2 x + 4 π ) ∣
there is long way to do the indefinite integral and a very short way.I will say shorter way.
I=(sin2x /2)/(sinx+cosx)
sinx+cosx = sqrt2 sin(pi/4 + x)
now sub pi/4+x=u
then integral becomes
cos2u/sinu forgeting constants(u should remeber to add it in the end)
I = integral 1-2sin^2u /sinu
I = integral cosecu-2sinu
I=log |cscu-cotu|+2cosu
i left out the constants.but take care of that and just put the limits.
There is a mistake @incredible mind
During substitution the numerator should be − cos 2 u not cos 2 u .
As sin ( 2 u − 2 π ) = − cos 2 u
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But i mentioned it already
cos2u/sinu 'forgeting constants'.......i consider constants irrelavemt in this integration...... .i am just trying to get the indefinite integral..i mentioned at the end "i left out the constants.but take care of that"..it is pain to type all that constants u know
Oh! I didn't know about this one. I did the long way! 2 pages long! Long way is by substituting t a n ( 2 x ) and partial fraction!
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And then the value of the expression asked turns out to be 1 9 .