Question

Find integration.
$\boxed{\tt \purple{\int \frac{\sin ^{6} x+\cos ^{6} x}{\sin ^{2} x \cos ^{2} x} d x}}$
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Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \frac{ {sin}^{6} x + {cos}^{6}x }{ {sin}^{2}x \: {cos}^{2}x} \: dx$

can be rewritten as

$\rm \:  =  \: \displaystyle\int\rm \frac{ {( {sin}^{2}x) }^{3} + ( { {cos}^{2}x) }^{3} }{ {sin}^{2}x \: {cos}^{2}x } \: dx$

We know,

$\boxed{\tt{ {x}^{3} + {y}^{3} = {(x + y)}^{3} - 3xy(x + y)}}$

So, using this identity, we get

$\rm \:  =  \: \displaystyle\int\rm \frac{ {({sin}^{2}x + {cos}^{2}x) }^{3} - 3 {sin}^{2}x{ {cos}^{2}x }( {sin}^{2} x + {cos}^{2}x)}{ {sin}^{2}x \: {cos}^{2}x } dx$

$\rm \:  =  \: \displaystyle\int\rm \frac{ 1 - 3 {sin}^{2}x{ {cos}^{2}x }}{ {sin}^{2}x \: {cos}^{2}x } dx$

$\rm \:  =  \: \displaystyle\int\rm \frac{1}{ {sin}^{2}x \: {cos}^{2}x}dx - 3\displaystyle\int\rm 1 \: dx$

$\rm \:  =  \: \displaystyle\int\rm \frac{ {sin}^{2} x + {cos}^{2} x}{ {sin}^{2}x \: {cos}^{2}x}dx - 3x + c$

$\rm \:  =  \: \displaystyle\int\rm \bigg(\frac{ {sin}^{2} x}{ {sin}^{2}x \: {cos}^{2}x} + \frac{ {cos}^{2}x }{ {sin}^{2}x {cos}^{2}x }\bigg) dx - 3x + c$

$\rm \:  =  \: \displaystyle\int\rm ( {sec}^{2}x + {cosec}^{2}x)dx - 3x + c$

$\rm \:  =  \: tanx - cotx - 3x + c$

$\purple{\boxed{\tt{ \displaystyle\int\rm \frac{ {sin}^{6} x + {cos}^{6}x }{ {sin}^{2}x \: {cos}^{2}x} \: dx = tanx - cotx - 3x + c}}}$

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MORE TO KNOW

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$