Question

cosecx(cosecx + cotx)dx​

Answer 1

Answer:

Step-by-step explanation:

I = ∫[cosecx(cosecx + cotx )].dx

I = ∫(x + cosecx . cotx).dx

I = ∫x .dx + ∫cosecx.cotx.dx

Using indentities:

∫cosec^2 x.dx= -cotx

∫cosecx.cotx.dx = -cosecx

I = -cotx - cosecx + C

Where C is Arbitrary Constant.

$ALWAY_SSMIL_E$  ^_^

Answer 2

Step-by-step explanation:

I= (x + cosecx . cotx) . dx

I= x.dx + cosecx. cotx. dx

identities are:

cosec^2x.cotx.dx= - cotx

cosecx.cotx.dx= - cosecx

I = - cotx - cosecx + c

c is the arbitrary constant....

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