Mathematics

# $\int \sqrt { 1 - \sin x } d x =$

$2\sqrt {1+\sin x}+C$

##### SOLUTION
$\\I=\int\sqrt{1-sinx}dx\\=\int\sqrt{1-sinx\times(\frac{1+sinx}{1+sinx})}dx$
$\\=\int\sqrt{(\frac{1^2-sin^2x}{1+sinx})}dx\\=\int(\frac{cosx}{\sqrt{1+sinx}})dx$
let 1+sinx=t
then cosx dx=dt
$\\or\>I=\int(\frac{1}{\sqrt t})dt$
$\\=(\frac{t^(\frac{1}{2})}{(\frac{1}{2})})+C\\2\sqrt{1+sinx}+C$

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

#### Realted Questions

Q1 Single Correct Medium
The value of $\displaystyle \int { \frac { d\left( { x }^{ 2 }+1 \right) }{ \sqrt { \left( { x }^{ 2 }+2 \right) } } }$ is equal to
• A. $\sqrt {(x^{2}-2)}+C$
• B. $2\sqrt {(x^{2}+3)}+C$
• C. $\sqrt {(x^{2}+2)}+C$
• D. $2\sqrt {(x^{2}+2)}+C$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
$\displaystyle \int { \left( \sqrt { { x }^{ 2 }+1 } \right) \left[ \frac { \ln { \left( { x }^{ 2 }+1 \right) } -2\ln { x } }{ { x }^{ 4 } } \right] dx } ;g\left( x \right)=\left( 1+\frac { 1 }{ { x }^{ 2 } } \right) ;$ equals
• A. $\displaystyle \frac { 1 }{ 3 } { \left( g\left( x \right) \right) }^{ 1/2 }\ln { \left( g\left( x \right) \right) } +\frac { 2 }{ 9 } \left( g\left( x \right) \right) +C$
• B. $\displaystyle \frac { 1 }{ 3 } { \left( g\left( x \right) \right) }^{ 3/2 }\ln { \left( g\left( x \right) \right) } -\frac { 2 }{ 9 } \left( g\left( x \right) \right) +C$
• C. None of these
• D. $\displaystyle -\frac { 1 }{ 3 } { \left( g\left( x \right) \right) }^{ 3/2 }\ln { \left( g\left( x \right) \right) } +\frac { 2 }{ 9 } { \left( g\left( x \right) \right) }^{ 3/2 }+C$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
$\int a^{3x+3}$ dx is equal to
• A. $\frac{a^{3x+3}}{log a}+c$
• B. $a^{3x+3}$ log a+c
• C. $3a^{3x+3}$ log a+c
• D. $\frac{a^{3x+3}}{3 log a}+c$

1 Verified Answer | Published on 17th 09, 2020

Q4 One Word Hard
If $\displaystyle I = \int \frac {x^2 - 2}{(x^4 + 5x^2 + 4) tan^{-1} (\displaystyle \frac {x^2 + 2}{x})} dx = \frac {A}{368} \log | tan^{-1} (x + 2/x) | + C$ then A equals.

$\int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx}$