Solve the integral ∫[π/4,cotx] (csc^2t)?
∫[π/4,cotx] (csc^2t)
∫[-π/4,3π/4] (9secxtanx)
Please explain how to solve these two integrals
Thank You :)
Update:The first integral is ∫(csc^2t) on the interval [π/4,cotx]
The second is ∫(9secxtanx) on the interval [-π/4,3π/4]
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Verified answer
∫dt/(sin t)² = -cot(t) + C
definite integral on [π/4,cot(x)] is
- cot(cot(x)) - (-cot(π/4)) = 1 - cot(cot(x))
for the second
substitute
cos x = u --> [-π/4,3π/4] becomes [1,-1]
- sin x dx = du
sin x dx = -du
∫9sec(x)tan(x) dx = 9∫sin(x)dx/(cos x)²=
=9∫-du/u² = 9/u + c
if we call a primitive
F(u) = 9/u
then the requested integral does not converge because
limit [F(-1) - F(𝛆) ] = ∞ (as 𝛆 ->0)
CLEAN IT UP!!!