Mathematics

# The value of $\int \dfrac {1}{\sqrt {\sin^{3}x \cos^{5}x}} dx$ is

$\dfrac {-2}{\sqrt {\tan x}} + \dfrac {2}{3} (\tan x)^{3/2} + C$

##### SOLUTION
Let $I=\int { \cfrac { 1 }{ \sqrt { \sin ^{ 3 }{ x } \cos ^{ 5 }{ x } } } dx }$
$I=\int { \cfrac { 1 }{ \sqrt { \cfrac { \sin ^{ 3 }{ x } }{ \cos ^{ 3 }{ x } } \cos ^{ 8 }{ x } } } dx } =\cfrac { 1 }{ \cos ^{ 4 }{ x } \sqrt { \tan ^{ 3 }{ x } } } dx$
$I=\int { \cfrac { \sec ^{ 4 }{ x } }{ \sqrt { \tan ^{ 3 }{ x } } } dx } =\int { \cfrac { \sec ^{ 3 }{ x } .\sec ^{ 2 }{ x } }{ \sqrt { \tan ^{ 3 }{ x } } } dx }$
$I=\int { \cfrac { \left( 1+\tan ^{ 2 }{ x } \right) }{ \sqrt { \tan ^{ 3 }{ x } } } \sec ^{ 2 }{ x } dx }$
Let $\tan { x } =t$
$\tan { x } =t$
$\tan { x } =t$
$\sec ^{ 2 }{ x } dx=dt$
So, $I=\int { \cfrac { 1+{ t }^{ 2 } }{ \sqrt { { t }^{ 3 } } } dt } =\int { \cfrac { 1+{ t }^{ 2 } }{ { t }^{ 3/2 } } dt }$
$I=\int { { t }^{ -3/2 }dt } +\int { { t }^{ 1/2 }dt }$
$I=\cfrac { { t }^{ -1/2 } }{ -1/2 } +\cfrac { { t }^{ 3/2 } }{ 3/2 } +c$
So, $I=\cfrac { -2 }{ \sqrt { \tan { x } } } +\cfrac { 2 }{ 3 } { \left( \tan { x } \right) }^{ 3/2 }+c$
where C is arbitary constant

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

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1 Verified Answer | Published on 17th 09, 2020

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On the basis of above information, answer the following questions :

Asked in: Mathematics - Limits and Derivatives

1 Verified Answer | Published on 17th 08, 2020