How do you integrate ∫3⋅(csc(t))2cot(t)dt?
Answers
First, note that because
3
is a constant, we can pull it out of the integral to simplify:
3
∫
csc
2
(
t
)
cot
(
t
)
d
t
Now - and this is the most important part - notice that the derivative of
cot
(
t
)
is
−
csc
2
(
t
)
. Because we have a function and its derivative present in the same integral, we can apply a
u
substitution like this:
u
=
cot
(
t
)
d
u
d
t
=
−
csc
2
(
t
)
d
u
=
−
csc
2
(
t
)
d
t
We can convert the positive
csc
2
(
t
)
to a negative like this:
−
3
∫
−
csc
2
(
t
)
cot
(
t
)
d
t
And apply the substitution:
−
3
∫
d
u
u
We know that
∫
d
u
u
=
ln
|
u
|
+
C
, so evaluating the integral is done. We just need to reverse substitute (put the answer back in terms of
t
) and attach that
−
3
to the result. Since
u
=
cot
(
t
)
, we can say:
−
3
(
ln
|
u
|
+
C
)
=
−
3
ln
|
cot
(
t
)
|
+
C
And that's all.
Answer: -6 cosec t +C
Step-by-step explanation:
As we can write it as
6 int cosec t . cot t dt
By formulae
Int cosecx . cotx dx= -cosecx+C
It will be...
=6 (-cosec t)
= -6 cosec t + C
Where C is the constant of integration...
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HOPE IT HELPS...