Question

Evaluate The Integral:-
$\colorbox{pink}{ \boxed{ \red{ \color{cyan} \bigstar \color{red}\displaystyle \tt \int {x}^{2} \sqrt{8 - {x}^{6} } \: \: \: \color{darkorange} \bigstar}}}$
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Answer 1

Evaluate The Integral:-

$\colorbox{pink}{ \boxed{ \red{ \color{cyan} \bigstar \color{red}\displaystyle \tt \int {x}^{2} \sqrt{8 - {x}^{6} } \: \: \: \color{darkorange} \bigstar}}}$

Answer with proper explanation.

Don't Dare for spam.

→ Expecting answer from :

★ Moderators

★ Brainly stars and teacher

★ Others best users

Answer 2

Question:-

Evaluate the integral:

$\sf \large \int {x}^{2} \sqrt{8 - {x}^{6} }.$

Given:-

$\sf \large \int {x}^{2} \sqrt{8 - {x}^{6} }.$

To Find:-

• Need to Evaluate the Given integral.

Solution:-

$\sf \large \int {x}^{2} \sqrt{8 - {x}^{6} }.$

$\sf \large Substitute \: \: u = x {}^{3} \longmapsto \frac{du}{dx} = 3x {}^{2} \\ \\ \sf \large dx = \frac{1}{3x {}^{2} } du, \: \\ \\ \sf \large use: \: x {}^{6} = u {}^{2} \\ \\ \sf \large = \frac{1}{3} \int \sqrt{8 - u {}^{2} } \: du$

Now Solving:

$\sf \large \int \sqrt{8 - u {}^{2} } \: du$

Perform Trigonometric substitution:

$\sf \large Substitute \: \: u = 2 {}^{ \frac{3}{2} } \: sin(v) \longmapsto \: v = arcsin \bigg( \frac{u}{2 {}^{ \frac{3}{2} } } \bigg) \\ \\ \sf \large du = 2 {}^{ \frac{3}{2} } \: cos(v) \: dv \\ \\ \sf \large = \int2 {}^{ \frac{3}{2} } \: cos(v) \sqrt{8 - 8 \: sin {}^{2}(v) } \: dv \\ \\ \sf \large Simplify \: using \: 8 - 8 \: sin {}^{2} (v) = 8 \: cos {}^{2} (v) : \\ \\ \sf \large = 8 \int cos {}^{2} (v)dv$

Now Solving:

$\sf \large \int cos {}^{2} (v)dv$

Apply reduction formula:

$\sf \large \int cos {}^{n}(v)dv = \frac{n - 1}{n} \int cos {}^{n - 2} (v)dv + \frac{cos {}^{n - 1} (v)sin(v)}{n} \\ \\ \sf \large with \: n = 2 : \\ \\ \sf \large = \frac{cos(v)sin(v)}{2} + \frac{1}{2 } \int 1 \: dv \\ \\ \sf \large or \: choose \: an \: alternative:$

Apply product - to - sum formulas

Now Solving:

$\sf \large \int \: 1 \: dv \\ \\ \sf \large Apply \: constant \: value: \\ \\ \sf \large = v$

Plug in solved integrals:

$\sf \large \frac{cos(v)sin(v)}{2} + \frac{1}{2} \int \: 1 \: dv \\ \\ \sf \large = \frac{cos(v)sin(v)}{2} + \frac{v}{2}$

Plug in solved integrals:

$\sf \large 8 \int \: cos {}^{2} (v)dv \\ \\ \sf \large = 4 \: cos(v)sin(v) + 4v$

$\sf \large Undo \: Substitution \: \: v = arcsin \bigg( \frac{u}{2 {}^{ \frac{3}{2} } } \bigg), \: use: \\ \\ \sf \large sin \bigg(arcsin \bigg( \frac{u}{2 {}^{ \frac{3}{2} } } \bigg) \bigg) = \frac{u}{2 {}^{ \frac{3}{2} } } \\ \\ \sf \large cos \bigg(arcsin \bigg( \frac{u}{2 {}^{ \frac{3}{2} } } \bigg) \bigg) = \sqrt{1 - \frac{u {}^{2} }{8} } \\ \\ \sf \large = 4 \: arcsin \bigg( \frac{u}{2 {}^{ \frac{3}{2} } } \bigg) + \sqrt{2} u \sqrt{1 - \frac{u {}^{2} }{8} }$

Plug in solved integrals:

$\sf \large \frac{1}{3} \int \sqrt{8 - u {}^{2} } \: du \\ \\ \sf \large = \frac{4 \: arcsin \bigg( \frac{u}{2 {}^{ \frac{3}{2} } } \bigg) }{3} + \frac{ \sqrt{2}u \sqrt{1 - \frac{u {}^{2} }{8} } }{3} \\ \\ \sf \large Undo \: Substitution \: \: u = x {}^{3} : \\ \\ \sf \large = \frac{4 \: arcsin \bigg( \frac{x {}^{3} }{2 {}^{ \frac{3}{2} } } \bigg)}{3} + \frac{ \sqrt{2}x {}^{3} \sqrt{1 - \frac{x {}^{6} }{8} } }{3}$

The problem is solved:

$\sf \large \int x {}^{2} \sqrt{8 - x {}^{6} } \: dx \\ \\ \sf \large = \frac{4 \: arcsin \bigg( \frac{x {}^{3} }{2 {}^{ \frac{3}{2} } } \bigg) }{3} + \frac{ \sqrt{2}x {}^{3} \sqrt{1 - \frac{x {}^{6} }{8} } }{ 3} + c$

$\sf \large Simplify: \: \: = \frac{8 \: arcsin \bigg( \frac{x {}^{3} }{2 {}^{ \frac{3}{2} } } \bigg) + x {}^{3} \sqrt{8 - x {}^{6} } }{6} + c$

Answer:-

$\sf \large \color{red}\sf \large \int {x}^{2} \sqrt{8 - {x}^{6} }= </p><p> \sf \large \frac{8 \: arcsin \bigg( \frac{x {}^{3} }{2 {}^{ \frac{3}{2} } } \bigg) + x {}^{3} \sqrt{8 - x {}^{6} } }{6} + c.$

Hope you have satisfied. ⚘