Question

reduction formula for cosec^n x​

Answer 1

Answer:

I

n

=∫csc

n

x=∫csc

n−2

xcsc

2

x

By integration by parts , we get I

n

=csc

n−2

x(−cotx)−∫(−cotx)(n−2)csc

n−3

x(−cscx)cotxdx=−csc

n−2

xcotx−(n−2)∫csc

n−2

xcot

2

x

I

n

=−csc

n−2

xcotx−(n−2)∫csc

n−2

x(csc

2

x−1)=

n−1

csc

n−2

xcotx

+

n−1

n−2

∫csc

n−2

x

Therefore the value of ∫csc

5

x=

4

csc

3

xcotx

+

4

3

∫csc

3

x

=

4

csc

3

xcotx

+

4

3

(

2

cscxcotx

+

2

1

∫cscx)

=

4

csc

3

xcotx

+

4

3cscxcotx

+

8

−3ln(cscx+cotx)

Step-by-step explanation:

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