Question

Hello ! Here's a calculus question!

Find that value of

$\\ \rm\Rrightarrow {\displaystyle{\int\limits_{\pi}^{2\pi}}}\dfrac{sin^6x+cos^6x}{sin^3xcos^3x}$


Note:-

Answer must include all steps properly .

Kindly don't waste time here if you don't know the answer .

All the best !​

Answer 1

Answer:

Undefined.

Formula's used:

$\longrightarrow \bold{sec(x) = \dfrac{1}{cos(x)} }$

$\longrightarrow \bold{cosec(x) = \dfrac{1}{sin(x)} }$

$\longrightarrow \bold{cot(x) = \dfrac{1}{tan(x)} }$

$\longrightarrow \bold{tan(x) = \dfrac{sinx}{cos(x)} }$

$\longrightarrow \bold{sin^2x + cos^2 x = 1 }$

$\longrightarrow \sf \bold{Sum \ Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx}$

$\longrightarrow \bold{ \int \:{sec^2(ax+b) = \dfrac{1}{a}tan(ax+b)+c }}$

$\longrightarrow \bold{\int \:\dfrac{1}{ax+b} =\dfrac{1}{a} ln|ax+b|+c}$

Explanation:

$\sf \Longrightarrow \int _{\pi }^{2\pi }\:\dfrac{sin^6(x)+cos^6(x)}{sin^3(x) \ * \ cos^3\left(x)}$

$\sf \Longrightarrow \sf \int _{\pi }^{2\pi }\:\dfrac{sin^6\left(x\right)}{sin^3\left(x\right)\cdot \:cos^3\left(x\right)} +\dfrac{cos^6\left(x\right)}{sin^3\left(x\right)\cdot \:cos^3\left(x\right)}$

$\Longrightarrow \sf \int _{\pi }^{2\pi }\:\dfrac{sin^3\left(x\right)}{\:cos^3\left(x\right)} +\dfrac{cos^3\left(x\right)}{sin^3\left(x\right)}$

                                                               

$\Longrightarrow \sf \int _{\pi }^{2\pi }\ tan^3(x)+ cot^3(x)$

$\sf \Longrightarrow \sf \int _{\pi }^{2\pi } \tan ^3(x)dx+\int _{\pi }^{2\pi } \cot ^3 (x)dx$

$\Longrightarrow \sf \bold{ [ }-\ln |\sec \left(x\right) |+\dfrac{\sec ^2(x)}{2}-\dfrac{\cot ^2 (x)}{2}-\ln |\sin(x)| \bold{ ] }^{2\pi }_\pi$

apply limits

$\sf \Longrightarrow \sf -\ln \left|\sec \left(2\pi \right)\right|+\dfrac{\sec ^2\left(2\pi\right)}{2}-\dfrac{\cot ^2\left(2\pi\right)}{2}-\ln \left|\sin \left(2\pi\right)\right|-(-\ln \left|\sec \left(\pi \right)\right|+\dfrac{\sec ^2\left(\pi\right)}{2}-\dfrac{\cot ^2\left(\pi\right)}{2}-\ln \left|\sin \left(\pi\right)\right|)$

simplify using trigonometric basic functions

$\Longrightarrow \sf -\ln \left|\dfrac{1}{\cos \left(2\pi \right)}\right|+\dfrac{\left(\dfrac{1}{\cos \left(2\pi \right)}\right)^2}{2}-\dfrac{\cot ^2\left(2\pi \right)}{2}-\ln \left|\sin \left(2\pi \right)\right|- ( -\ln \left|\dfrac{1}{\cos \left(\pi \right)}\right|+\dfrac{\left(\frac{1}{\cos \left(\pi \right)}\right)^2}{2}-\dfrac{\cot ^2\left(\pi \right)}{2}-\ln \left|\sin \left(\pi \right)\right|)$

• The value of cot(2π) is not defined. [ cot(2π) = ∞ ]

⇒  Undefined

Answer 2

Answer:

$\displaystyle \int\limits^{2 \pi}_{\pi} {\frac{\sin^6 x + \cos^6 x}{\sin^3 x \cos^3 x}} \, dx = \boxed{\text{un} \text{de} \text{f}}$

General Formulas and Concepts:
Calculus

Differentiation

• Derivatives
• Derivative Notation

Integration

• Integrals

Integration Rule [Reverse Power Rule]:
$\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

Integration Rule [Fundamental Theorem of Calculus 1]:
$\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]:
$\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Integration Property [Addition/Subtraction]:
$\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

Integration Methods: U-Substitution + U-Solve

Step-by-step explanation:

Step 1: Define

Identify given.

$\displaystyle \int\limits^{2 \pi}_{\pi} {\frac{\sin^6 x + \cos^6 x}{\sin^3 x \cos^3 x}} \, dx$

Step 2: Integrate Pt. 1

1. [Integrand] Rewrite:
   $\displaystyle \frac{\sin^6 x + \cos^6 x}{\sin^3 x \cos^3 x} = \frac{\sin^6 x}{\sin^3 x \cos^3 x} + \frac{\cos^6 x}{\sin^3 x \cos^3 x}$
2. [Integrand] Simplify:
   $\displaystyle \frac{\sin^6 x + \cos^6 x}{\sin^3 x \cos^3 x} = \tan^3 x + \cot^3 x$
3. [Integrand] Rewrite:
   $\displaystyle \frac{\sin^6 x + \cos^6 x}{\sin^3 x \cos^3 x} = \tan^2 x (\sec^2 x - 1) + \cot^2 x (\csc^2 x - 1)$

Step 3: Integrate Pt. 2

1. [Integral] Rewrite:
   $\displaystyle \int\limits^{2 \pi}_{\pi} {\frac{\sin^6 x + \cos^6 x}{\sin^3 x \cos^3 x}} \, dx = \int\limits^{2 \pi}_{\pi} {\tan^2 x (\sec^2 x - 1) + \cot^2 x (\csc^2 x - 1)} \, dx$
2. [Integral] Rewrite [Integration Property - Addition/Subtraction]:
   $\displaystyle \int\limits^{2 \pi}_{\pi} {\frac{\sin^6 x + \cos^6 x}{\sin^3 x \cos^3 x}} \, dx = \int\limits^{2 \pi}_{\pi} {\tan^2 x (\sec^2 x - 1)} \, dx + \int\limits^{2 \pi}_{\pi} {\cot^2 x (\csc^2 x - 1)} \, dx$

Step 4: Integrate Pt. 3

Identify variables for u-solve.

1st Integral

1. Set u:
   $\displaystyle u = \sec x$
2. [u] Apply Trigonometric Differentiation:
   $\displaystyle du = \sec x \tan x \, dx$
3. [du] Rewrite:
   $\displaystyle dx = \frac{1}{\sec x \tan x} \, du$

2nd Integral

1. Set v:
   $\displaystyle v = \csc x$
2. [v] Apply Trigonometric Differentiation:
   $\displaystyle dv = - \cot x \csc x \, dx$
3. [dv] Rewrite:
   $\displaystyle dx = \frac{-1}{\cot x \csc x} \, dv$

Step 5: Integrate Pt. 4

1. [Integrals] Apply Integration Method [U-Solve]:
   $\displaystyle \int\limits^{2 \pi}_{\pi} {\frac{\sin^6 x + \cos^6 x}{\sin^3 x \cos^3 x}} \, dx = \int\limits^{x = 2 \pi}_{x = \pi} {\frac{\tan^2 x (\sec^2 x - 1)}{\sec x \tan x}} \, du + \int\limits^{x = 2 \pi}_{x = \pi} {\frac{- \cot^2 x (\csc^2 x - 1)}{\cot x \csc x}} \, dv$
2. [Integrals] Simplify:
   $\displaystyle \int\limits^{2 \pi}_{\pi} {\frac{\sin^6 x + \cos^6 x}{\sin^3 x \cos^3 x}} \, dx = \int\limits^{x = 2 \pi}_{x = \pi} {\frac{u^2 - 1}{u}} \, du - \int\limits^{x = 2 \pi}_{x = \pi} {\frac{v^2 - 1}{v}} \, dv$
3. [Integrals] Apply Integration Rule [Reverse Power Rule]:
   $\displaystyle \int\limits^{2 \pi}_{\pi} {\frac{\sin^6 x + \cos^6 x}{\sin^3 x \cos^3 x}} \, dx = \frac{u^2}{2} \bigg| \limits^{x = 2 \pi}_{x = \pi} - \int\limits^{x = 2 \pi}_{x = \pi} {\frac{1}{u}} \, du - \Bigg( \frac{v^2}{2} \bigg| \limits^{x = 2 \pi}_{x = \pi} - \int\limits^{x = 2 \pi}_{x = \pi} {\frac{1}{v}} \, dv \Bigg)$
4. [Integrals] Apply Logarithmic Integration:
   $\displaystyle \int\limits^{2 \pi}_{\pi} {\frac{\sin^6 x + \cos^6 x}{\sin^3 x \cos^3 x}} \, dx = \frac{u^2}{2} \bigg| \limits^{x = 2 \pi}_{x = \pi} - \ln | u | \bigg| \limits^{x = 2 \pi}_{x = \pi} - \Bigg( \frac{v^2}{2} \bigg| \limits^{x = 2 \pi}_{x = \pi} - \ln | v | \bigg| \limits^{x = 2 \pi}_{x = \pi} \Bigg)$
5. Back-Substitute variables u and v:
   $\displaystyle \int\limits^{2 \pi}_{\pi} {\frac{\sin^6 x + \cos^6 x}{\sin^3 x \cos^3 x}} \, dx = \frac{\sec^2 x}{2} \bigg| \limits^{2 \pi}_{\pi} - \ln | \sec x | \bigg| \limits^{2 \pi}_{\pi} - \Bigg( \frac{\csc^2 x}{2} \bigg| \limits^{2 \pi}_{\pi} - \ln | \csc x | \bigg| \limits^{2 \pi}_{\pi} \Bigg)$
6. Simplify:
   $\displaystyle \int\limits^{2 \pi}_{\pi} {\frac{\sin^6 x + \cos^6 x}{\sin^3 x \cos^3 x}} \, dx = \bigg( \ln | \csc x | - \ln | \sec x | + \frac{\sec^2 x - \csc^2 x}{2} \bigg) \bigg| \limits^{2 \pi}_{\pi}$
7. Apply Integration Rule [Fundamental Theorem of Calculus 1]:
   $\displaystyle \int\limits^{2 \pi}_{\pi} {\frac{\sin^6 x + \cos^6 x}{\sin^3 x \cos^3 x}} \, dx = \boxed{\text{un} \text{de} \text{f}}$

∴ we have found the value of the given integral.

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Learn more about integration: https://brainly.com/question/14413972

Learn more about calculus: https://brainly.com/question/20197752

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Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration