Mathematics

Solve : $$\dfrac{dx}{\sin^3x .\cos^5x}$$


SOLUTION
$$I=\int { \dfrac { dx }{ \left( { \ sin }^{ 3 }x.{ \ cos }^{ 5 }x \right)  }  } \\$$ substituting,\\ $$\sin x=\dfrac { \tan x }{ \sec x } \\ \cos x\dfrac { 1 }{ \sec x } \\ { \sec }^{ 2 }x={ \tan }^{ 2 }x+1\\$$ $$\therefore I=\int { { \sec }^{ 2 }x } .\dfrac { \left( \tan^{ 2 }x+1 \right) ^{ 3 } }{ { \tan }^{ 3 }x } dx\\ Let,\\ \quad \quad u=\tan x\\$$ $$\rightarrow dx=\dfrac { 1 }{ { \sec }^{ 2 } } du\\$$ $$\therefore I=\int { \dfrac { { \left( { u }^{ 2 }+1 \right)  }^{ 3 } }{ { u }^{ 3 } } du } \\ \quad \quad =\int { \left( { u }^{ 3 }+3u+\dfrac { 3 }{ u } +\dfrac { 1 }{ { u }^{ 3 } }  \right) du }$$ \\ $$\quad \quad =\int { { u }^{ 3 }du } +3\int { udu } +3\int { \dfrac { 1 }{ u } du } +\int { \dfrac { 1 }{ { u }^{ 3 } } du }$$ \\ $$\quad \quad =3\log u+\dfrac { { u }^{ 4 } }{ 4 } +\dfrac { { 3u }^{ 2 } }{ 2 } -\dfrac { 1 }{ { 2u }^{ 2 } }$$
=$$3 \log \ tanx$$$$+ \dfrac { { \tan }^{ 4 }x }{ 4 } +\dfrac { 3{ \tan }^{ 2 }x }{ 2 } -\dfrac { 1 }{ 2{ tan }^{ 2 }x } +C$$
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Subjective Medium Published on 17th 09, 2020
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