Question: "Evaluate the integral. ∫csc44xdx"

High School
Calculus and Analysis
Calculus 2
Applications of integrals

elvishwitchxyp

Answered question

2021-12-13

Evaluate the integral.

\int \csc^{4} 4 x d x

Answer & Explanation

Becky Harrison

Beginner2021-12-14Added 40 answers

Step 1
Consider the following integral:

\int \csc^{4} (4 x) d x

Substitute

4 x = u \Rightarrow 4 d x = d u \Rightarrow d x = \frac{1}{4} d u

in the above integral:

\int \csc^{4} (4 x) d x = \frac{1}{4} \int \csc^{4} (u) d u

= \frac{1}{4} \int [\cot^{2} (u) + 1] \csc^{2} (u) d u

Step 2
Substitute

\cot (u) = v \Rightarrow - \csc^{2} (u) d u = d v \Rightarrow \csc^{2} (u) d u = - d v

in the above equation:

\int \csc^{4} (4 x) d x = - \frac{1}{4} \int (v^{2} + 1) d v

= - \frac{1}{4} (\frac{v^{3}}{3} + v) + c

Substitute

v = \cot (u)

in the above equation:

\int \csc^{4} (4 x) d x = - \frac{1}{4} [\frac{\cot^{3} (u)}{3} + \cot (u)] + c

Substitute u=4x in the above equation:

\int \csc^{4} (4 x) d x = - \frac{1}{4} [\frac{\cot^{3} (4 x)}{3} + \cot (4 x)] + c

Step 3
Hence, the solution is

- \frac{\cot^{3} (4 x)}{12} - \frac{\cot (4 x)}{4} + c

vicki331g8

Beginner2021-12-15Added 37 answers

Given:
$\int {\csc (4 x)}^{4} d x$
$\int \frac{\csc (t)^{4}}{4} d t \frac{1}{4} \times \int \csc (t)^{4} d t$
$\frac{1}{4} \times (- \frac{1}{3} \times {\csc (t)}^{2} \cot (t) + \frac{2}{3} \times \int {\csc (t)}^{2} d t)$
$\frac{1}{4} \times (- \frac{1}{3} \times {\csc (t)}^{2} \cot (t) + \frac{2}{3} \times (- \cot (t)))$
$\frac{1}{4} \times (- \frac{1}{3} \times {\csc (4 x)}^{2} \cot (4 x) + \frac{2}{3} \times (- \cot (4 x)))$
Simplify
$- \frac{\cos (4 x)}{12 {\sin (4 x)}^{3}} - \frac{\cos (4 x)}{6 \sin (4 x)}$
Add C
Answer:
$- \frac{\cos (4 x)}{12 {\sin (4 x)}^{3}} - \frac{\cos (4 x)}{6 \sin (4 x)} + C$

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