Evaluate the integral: ∫csc((v-π)/2)cot((v-π)/2)

[cookiemnstr510510]

Homework Statement

Evaluate the integral:
∫csc((v-π)/2)cot((v-π)/2)

Homework Equations

cscx=1/sinx
cot=sinx/cosx

The Attempt at a Solution

I first turned csc and cot into the above "relevant equations"
∫ (1/sin($\frac{v-π}{2}$)($\frac{sin(v-π)/2}{cos(v-π)/2}$)=∫cos^-1((v-π)/2)
then
U=(v-π)/2
dU=(1/2)dv
dv=2dU
→2∫cos^-1(u)dU= [-2/(√1-((v-π)/2)]+C

is there another way to do this? I feel like i may have done it wrong.

Thanks and Merry Christmas!

 

[scottdave]

You have a mistake. You replaced 1/cos with cos^-1, which actually means the inverse cosine. Then your answer is the integral of inverse cosine, not 1/cosine

 

[LCKurtz]

Homework Statement

Evaluate the integral:
∫csc((v-π)/2)cot((v-π)/2)

Homework Equations

cscx=1/sinx
cot=sinx/cosx

The Attempt at a Solution

I first turned csc and cot into the above "relevant equations"
∫ (1/sin($\frac{v-π}{2}$)($\frac{sin(v-π)/2}{cos(v-π)/2}$)=∫cos^-1((v-π)/2)
then
U=(v-π)/2
dU=(1/2)dv
dv=2dU
→2∫cos^-1(u)dU= [-2/(√1-((v-π)/2)]+C

is there another way to do this? I feel like i may have done it wrong.

Thanks and Merry Christmas!

There is no reason to switch to sines and cosines. Just remember the relationships between cosecants and cotangents and their derivatives. Try the same type of substitution directly with the cosecant.

 

[SammyS]

Homework Statement

Evaluate the integral:
∫csc((v-π)/2)cot((v-π)/2)

Homework Equations

cscx=1/sinx
cot=sinx/cosx
...

You are missing the variable of integration. Integration should be of the form: ∫ ... dv .

Also, you have an incorrect identity for cot(x) .