Mathematics

# $\displaystyle\int {{{dx} \over {\sqrt {{{\sin }^3}x\cos x} }} = g(x)} + c \Rightarrow g(x) =$

$\displaystyle{{ - 2} \over {\sqrt {\tan x} }}$

##### SOLUTION
Solution:
$I=\displaystyle \int { \dfrac{ 1 }{ \sqrt { \sin ^{ 3 }{ x } \cos { x } } } dx }$

$=\displaystyle \int { \dfrac{ 1 }{ \dfrac{ \sqrt { \sin ^{ 3 }{ x } } }{ \sqrt { \cos ^{ 3 }{ x } } } \times \sqrt { \cos ^{ 3 }{ x } } \times \sqrt { \cos { x } } } dx }$

$=\displaystyle \int { \dfrac{ \sec ^{ 2 }{ x } .dx }{ \sqrt { \tan ^{ 3 }{ x } } } }$      Let $\tan x=t$

$\Rightarrow \sec^2x dx=dt$

$=\displaystyle \int { \dfrac{ dt }{ { t }^{ 3/2 } } }$

$=-2t^{-1/2}+C$

$=\dfrac {-2}{\sqrt {\tan x}}+C=g(x)+C$

$\Rightarrow \ g(x)=\dfrac {-2}{\sqrt {\tan x}}$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Single Correct Hard
$\displaystyle \int { { e }^{ x }\frac { \left( { x }^{ 2 }+1 \right) }{ { \left( x+1 \right) }^{ 2 } } dx } =$
• A. $\displaystyle \left( \frac { x+1 }{ x-1 } \right) { e }^{ x }+c$
• B. ${ e }^{ x }\left( x+1 \right) \left( x-1 \right) +c$
• C. none of these
• D. $\displaystyle \left( \frac { x-1 }{ x+1 } \right) { e }^{ x }+c$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
Let $F(x)=f(x)+f\left ( \dfrac{1}{x} \right )$, where $f(x)=\int_{l}^{x}\dfrac{logt}{l+t}dt$. Then $F(e)$ equals
• A.
• B. 1
• C. 2
• D. $\dfrac{1}{2}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
The value of $\displaystyle\int { \cfrac { 2 }{ { \left( { e }^{ x }+{ e }^{ -x } \right) }^{ 2 } } } dx$ is
• A. $-\cfrac { 1 }{ { e }^{ x }+{ e }^{ -x } } +C$
• B. $\cfrac { -1 }{ { \left( { e }^{ x }+1 \right) }^{ 2 } } +C$
• C. $\cfrac { 1 }{ { e }^{ x }-{ e }^{ -x } } +C\quad \quad$
• D. $\cfrac { -{ e }^{ -x } }{ { e }^{ x }+{ e }^{ -x } } +C\quad$

1 Verified Answer | Published on 17th 09, 2020

Q4 Multiple Correct Medium
$\displaystyle \int \frac{dx}{x^{3}\left ( 1 - \displaystyle \frac{1}{2x^{2}} \right )}$ equals
• A. $ln| 2x^{2} - 1| + 2\, ln |x| + C$
• B. $ln| 2x^{2} - 1| - 2\, ln |x| + C$
• C. $ln| 2x^{2} - 1| - ln (x^{2}) - ln2 + C$
• D. $ln \left | 1 - \displaystyle \frac{1}{2x^{2}} \right | + c$

Given that for each $\displaystyle a \in (0, 1), \lim_{h \rightarrow 0^+} \int_h^{1-h} t^{-a} (1 -t)^{a-1}dt$ exists. Let this limit be $g(a)$
In addition, it is given that the function $g(a)$ is differentiable on $(0, 1)$