(1 + tanx)/sinx
Multiply by sinx/sinx
sinx + tanxsinx
Divide by sin2x (1/sin2x) = cscx
cscx + tan(x)csc(x)
tanx = sinx/cosx and cscx = 1/sinx
cscx + (sinx/cosx)(1/sinx)
sinx cancels out
cscx + 1/cosx
1/cosx = secx
cscx + secx
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pi divided by 6 is a constant and so its first derivative is 0. And since that is also a constant, the second derivative is 0. It is not clear what f(x) = csc(x) has to do with that!
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sinx cscx = 1 is the same thing as sinx(1/sinx) = 1 which is the same as sinx/sinx = 1. This evaluates to 1=1, which is true.
According to Wolfram Alpha, input:integral csc x it is -log[cot(x) + csc(x)] + constant You can verify this by taking the derivative of the purported integral.
Trig functions have their own special derivatives that you will have to memorize. For instance: the derivative of sinx is cosx. The derivative of cosx is -sinx The derivative of tanx is sec2x The derivative of cscx is -cscxcotx The derivative of secx is secxtanx The derivative of cotx is -csc2x