Question

#Quality Question
@Integration

Find Value of
$\int {sec} \: ^ {2/3} x \: cosec \: ^{4/3} x \: dx$


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Answer 1

SOLUTION-;

$reffered \: to \: the \: attachment$

Answer 2

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

$\rm :\longmapsto\:\displaystyle\int\rm {sec} \: ^ {2/3} x \: cosec \: ^{4/3} x \: dx$

We know,

$\boxed{ \sf{ \:secx = \frac{1}{cosx}}} \: \: and \: \: \boxed{ \sf{ \:cosecx = \frac{1}{sinx}}}$

So, using this

$\rm \:  =  \:  \:\displaystyle\int\rm \dfrac{ dx }{{\bigg(cosx\bigg) }^{\dfrac{2}{3} }{\bigg(sinx\bigg) }^{\dfrac{4}{3} }}$

can be rewritten as

$\rm \:  =  \:  \:\displaystyle\int\rm \dfrac{ dx }{{\bigg(cosx\bigg) }^{\dfrac{2}{3} }{\bigg(cosx \times \dfrac{sinx}{cosx} \bigg) }^{\dfrac{4}{3} }}$

$\rm \:  =  \:  \:\displaystyle\int\rm \dfrac{ dx }{{\bigg(cosx\bigg) }^{\dfrac{2}{3} }{\bigg(cosx \times tanx \bigg) }^{\dfrac{4}{3} }}$

We know,

$\boxed{ \sf{ \: {(xy)}^{m} = {x}^{m} \times {y}^{m}}}$

So using this, we get

$\rm \:  =  \:  \:\displaystyle\int\rm \dfrac{ dx }{{\bigg(cosx\bigg) }^{\dfrac{2}{3} }{\bigg(cosx\bigg) }^{\dfrac{4}{3} }{\bigg(tanx \bigg) }^{\dfrac{4}{3} }}$

$\rm \:  =  \:  \:\displaystyle\int\rm \dfrac{ dx }{{\bigg(cosx\bigg) }^{\dfrac{2}{3} + \dfrac{4}{3} }{\bigg(tanx \bigg) }^{\dfrac{4}{3} }}$

$\rm \:  =  \:  \:\displaystyle\int\rm \dfrac{ dx }{{\bigg(cosx\bigg) }^{\dfrac{2 + 4}{3}}{\bigg(tanx \bigg) }^{\dfrac{4}{3} }}$

$\rm \:  =  \:  \:\displaystyle\int\rm \dfrac{ dx }{{\bigg(cosx\bigg) }^{\dfrac{6}{3}}{\bigg(tanx \bigg) }^{\dfrac{4}{3} }}$

$\rm \:  =  \:  \:\displaystyle\int\rm \dfrac{ dx }{{\bigg(cosx\bigg) }^{2}{\bigg(tanx \bigg) }^{\dfrac{4}{3} }}$

$\rm \:  =  \:  \:\displaystyle\int\rm \dfrac{ {sec}^{2}xdx }{{\bigg(tanx \bigg) }^{\dfrac{4}{3} }}$

Now,

To evaluate this integral, we use method of Substitution.

$\red{\rm :\longmapsto\:Put \: tanx = y} \\ \red{\rm :\longmapsto\: {sec}^{2}xdx = dy}$

So, above integral can be rewritten as

$\rm \:  =  \:  \:\displaystyle\int\rm \dfrac{dy}{{\bigg(y\bigg) }^{ \dfrac{4}{3} }}$

$\rm \:  =  \:  \:\displaystyle\int\rm {\bigg(y\bigg) }^{ - \dfrac{4}{3} } \: dy$

We know,

$\boxed{ \sf{ \:\displaystyle\int\rm {x}^{n} \: dx \: = \: \frac{ {x}^{n + 1} }{n + 1} \: + \: c}}$

So, using this result, we get

$\rm \:  =  \:  \:\dfrac{{\bigg(y\bigg) }^{ - \dfrac{4}{3} + 1}}{ - \dfrac{4}{3} + 1} + c$

$\rm \:  =  \:  \:\dfrac{{\bigg(y\bigg) }^{ - \dfrac{1}{3}}}{ - \dfrac{1}{3}} + c$

$\rm \:  =  \:  \: - 3{\bigg(y\bigg) }^{ - \dfrac{1}{3} } + c$

$\rm \:  =  \:  \: - 3{\bigg(tanx\bigg) }^{ - \dfrac{1}{3} } + c$

$\rm \:  =  \:  \: - 3{\bigg(cotx\bigg) }^{ \dfrac{1}{3} } + c$

$\rm \:  =  \:  \: - 3 \: \sqrt[3]{cotx} \: + \: c$

Additional Information :-

$\boxed{ \sf{ \:\displaystyle\int\rm k \: dx = kx + c}}$

$\boxed{ \sf{ \:\displaystyle\int\rm cosx \: dx = sinx + c}}$

$\boxed{ \sf{ \:\displaystyle\int\rm sinx \: dx = - \: cosx + c}}$

$\boxed{ \sf{ \:\displaystyle\int\rm tanx \: dx = - \: log \: cosx + c}}$

$\boxed{ \sf{ \:\displaystyle\int\rm cotx \: dx = \: log \: sinx + c}}$

$\boxed{ \sf{ \:\displaystyle\int\rm secx \: dx = \: log \: (secx \: + \: tanx) + c}}$

$\boxed{ \sf{ \:\displaystyle\int\rm cosecx \: dx = \: log \: (cosecx \: - \: cotx) + c}}$

$\boxed{ \sf{ \:\displaystyle\int\rm {sec}^{2}x = tanx + c}}$

$\boxed{ \sf{ \:\displaystyle\int\rm {cosec}^{2}x = - \: cotx + c}}$