Solve the trigonometric integral $\int\csc\left(x\right)^3dx$

Step-by-step Solution

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Final answer to the problem

$-\frac{1}{2}\cot\left(x\right)\csc\left(x\right)-\frac{1}{2}\ln\left|\csc\left(x\right)+\cot\left(x\right)\right|+C_0$
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Step-by-step Solution

How should I solve this problem?

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  • Integrate by partial fractions
  • Integrate by substitution
  • Integrate by parts
  • Integrate using tabular integration
  • Integrate by trigonometric substitution
  • Weierstrass Substitution
  • Integrate using trigonometric identities
  • Integrate using basic integrals
  • Product of Binomials with Common Term
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1

Rewrite the trigonometric function $\csc\left(x\right)^3$ as the product of two lower exponents

$\int\csc\left(x\right)^2\csc\left(x\right)dx$

Learn how to solve limits by direct substitution problems step by step online.

$\int\csc\left(x\right)^2\csc\left(x\right)dx$

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Learn how to solve limits by direct substitution problems step by step online. Solve the trigonometric integral int(csc(x)^3)dx. Rewrite the trigonometric function \csc\left(x\right)^3 as the product of two lower exponents. We can solve the integral \int\csc\left(x\right)^2\csc\left(x\right)dx by applying integration by parts method to calculate the integral of the product of two functions, using the following formula. First, identify or choose u and calculate it's derivative, du. Now, identify dv and calculate v.

Final answer to the problem

$-\frac{1}{2}\cot\left(x\right)\csc\left(x\right)-\frac{1}{2}\ln\left|\csc\left(x\right)+\cot\left(x\right)\right|+C_0$

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Function Plot

Plotting: $-\frac{1}{2}\cot\left(x\right)\csc\left(x\right)-\frac{1}{2}\ln\left(\csc\left(x\right)+\cot\left(x\right)\right)+C_0$

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7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

How to improve your answer:

Main Topic: Limits by Direct Substitution

Find limits of functions at a specific point by directly plugging the value into the function.

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