Mathematics

# Evaluate $\int {\dfrac{{{{\cos }^2}x}}{{{{\sin }^3}x{{\left( {{{\sin }^5}x + {{\cos }^5}x} \right)}^{\frac{3}{5}}}}}dx}$

##### SOLUTION
$I=\int { \dfrac { { { { \cos }^{ 4 } }x } }{ { { { \sin }^{ 3 } }x{ { \left( { { { \sin }^{ 5 } }x+{ { \cos }^{ 5 } }x } \right) }^{ \frac { 3 }{ 5 } } } } } dx }$
$I =\int { \dfrac { { { { \sec }^{ 2 } }xdx } }{ { { { \tan }^{ 6 } }x{ { \left( { 1+\frac { 1 }{ { { { \tan }^{ 5 } }x } } } \right) }^{ \frac { 3 }{ 5 } } } } } }$

Let
$\tan x=p\, \, \, \, { \sec ^{ 2 } }xdx=dp$

Therefore,
$\int { \dfrac { { dp } }{ { { p^{ 6 } }{ { \left( { 1+\frac { 1 }{ { { p^{ 5 } } } } } \right) }^{ \frac { 3 }{ 5 } } } } } } \\ 1+\dfrac { 1 }{ { { p^{ 5 } } } } =k \\ \dfrac { { dk } }{ { dp } } =\dfrac { { -5 } }{ { { p^{ 6 } } } } \\$

Therefore,
$I=-\dfrac { { 1 } }{ 5 } \int { { { \left( k \right) }^{ \frac { { -3 } }{ 5 } } }dk } \\ =\dfrac { { -1 } }{ 5 } \left( { \dfrac { 5 }{ 2 } } \right) { k^{ \frac { 2 }{ 5 } } }+C \\ =\dfrac { { -1 } }{ 2 } { \left[ { \dfrac { { { p^{ 5 } }+1 } }{ { { p^{ 5 } } } } } \right] ^{ \frac { 2 }{ 5 } } } \\ =-\dfrac { 1 }{ 2 } { \left[ { \dfrac { { { { \tan }^{ 5 } }x+1 } }{ { { { \tan }^{ 5 } }x } } } \right] ^{ \frac { 2 }{ 5 } } }+C \\ =-\dfrac { 1 }{ 2 } { \left( { 1+{ { \cot }^{ 5 } }x } \right) ^{ \frac { 2 }{ 5 } } }+C$

Hence, this is the answer.

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Subjective Medium Published on 17th 09, 2020
Questions 203525
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