Find out the answer to "Integral Calculus: Trigonometry and Inverse Trigonometry. Evaluate the..."

vegetars8t

Answered question

2021-12-11

Integral Calculus: Trigonometry and Inverse Trigonometry. Evaluate the following

\int \csc^{6} u d u

Answer & Explanation

SlabydouluS62

Skilled2021-12-12Added 52 answers

Given
The trigonometric function is $\int \cos e c^{6} u d u$ .
Evaluate
$\int \cos e c^{6} u d u = \int \cos e c^{4} u \cdot \int \cos e c^{2} u d u$
$\int {(\cos e c^{2} u)}^{2} \cdot \cos e c^{2} u d u$
$\int {(1 + \cot^{2} u)}^{2} \cdot \cos e c^{2} u d u$
(Since, $\cos e c^{2} u = 1 + \cot^{2} u$ )
Let $t = \cot u$
$\frac{d t}{d u} = - \cos e c^{2} u$ implies $- d t = \cos e c^{2} u d u$
Now, $\int \cos e c^{6} u d u = - \int (1 + t^{2})^{2} d t$
$= - \int (1 + 2 t^{2} + t^{4}) d t$
$= - (t + \frac{2 t^{3}}{3} + \frac{t^{5}}{5}) + c$
$= - t - \frac{2 t^{3}}{3} - \frac{t^{5}}{5} + c$
Here $t = \cot u$
Therefore, $\int \cos e c^{6} u d u = - \cot u - \frac{2}{3} \cot^{3} u - \frac{1}{5} c$

Jack Maxson

Beginner2021-12-13Added 25 answers

I looked at the solutions on this site and was able to solve theZSK that I did not understand for a long time, thank you very much!

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