Mathematics

# If $\displaystyle\int {\dfrac{1}{{1 + \sin x}}dx = \tan \left( {\frac{x}{2} + a} \right) + b}$, then

$a = - \dfrac{\pi }{4},b \in R$

##### SOLUTION
$\displaystyle \int \dfrac {dx}{1+\sin x}=\tan \left (\dfrac {x}{2}+a \right)+b$
$=\displaystyle \int \dfrac {dx}{1+\sin x}\dfrac {(1-\sin x)}{(1-\sin x)}=\displaystyle \int \dfrac {(1-\sin x)}{\cos^2 x}dx$
$=\displaystyle \int (\sec^2 x-\tan x \sec x)dx$
$=\tan x-\sec x-c=$
$=\dfrac { \sin { x } -1 }{ \cos { x } } +c\dfrac { 2\cos { \dfrac { x }{ 2 } \sin { \dfrac { x }{ 2 } -1 } } }{ \cos ^{ 2 }{ \left( \dfrac { x }{ 2 } \right) } -\sin ^{ 2 }{ \left( \dfrac { x }{ 2 } \right) } }$
$=\dfrac { 2\cos { \dfrac { x }{ 2 } } \sin { \dfrac { x }{ 2 } } -\left( \sin ^{ 2 }{ \dfrac { x }{ 2 } } +\cos ^{ 2 }{ \dfrac { x }{ 2 } } \right) }{ \left( \cos { \dfrac { x }{ 2 } } -\sin { \dfrac { x }{ 2 } } \right) \left( \cos { \dfrac { x }{ 2 } } +\sin { \dfrac { x }{ 2 } } \right) } =-\dfrac { \left( \cos { \dfrac { x }{ 2 } } -\sin { \dfrac { x }{ 2 } } \right) }{ \left( \cos { \dfrac { x }{ 2 } } +\sin { \dfrac { x }{ 2 } } \right) }$
$=\dfrac { -\left( \dfrac { 1 }{ \sqrt { 2 } } \cos { \dfrac { x }{ 2 } } +\dfrac { 1 }{ \sqrt { 2 } } \sin { \dfrac { x }{ 2 } } \right) =\sin { \left( \dfrac { x }{ 2 } -\dfrac { \pi }{ 4 } \right) } }{ -\left( \dfrac { 1 }{ \sqrt { 2 } } \cos { \dfrac { x }{ 2 } } -\dfrac { 1 }{ \sqrt { 2 } } \sin { \dfrac { x }{ 2 } } \right) =\sin { \left( \dfrac { x }{ 2 } -\dfrac { \pi }{ 4 } \right) } } =\tan { \left( \dfrac { x }{ 2 } -\dfrac { \pi }{ 4 } \right) }$
$a=\dfrac { \pi }{ 4 } ,b\in R$

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

#### Realted Questions

Q1 Subjective Medium
Evaluate the following integral:
$\int { \tan ^{ 3/2 }{ x } \sec ^{ 2 }{ x } } dx\quad$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
Consider the integral $I=\displaystyle\int^{\pi}_0 ln(\sin x)dx$.What is $\displaystyle\int^{\dfrac{\pi}{2}}_{0}$ ln $(\sin x)dx$ equal to?
• A. $4I$
• B. $2I$
• C. $I$
• D. $\dfrac{I}{2}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
The value of $\displaystyle \int_{1}^{2}\left [ f\left \{ g\left ( x \right ) \right \} \right ]^{-1}.{f}'\left \{ g\left ( x \right ) \right \}.{g}'\left ( x \right )dx,$ where $\displaystyle g\left ( 1 \right )=g\left ( 2 \right ),$ is equal to?
• A. $1$
• B. $2$
• C. none of these
• D. $0$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$\displaystyle \int\frac{(x+1)}{x(1+xe^{x})}dx=$
• A. $\displaystyle \log|\frac{e^{x}}{1+xe^{x}}|+c$
• B. $-\displaystyle \log|\frac{e^{x}}{1+xe^{x}}|+c$
• C. $-\log|\displaystyle \frac{e^{x}}{1-xe^{x}}|+c$
• D. $\displaystyle \log|\frac{xe^{x}}{1+xe^{x}}|+c$

Consider two differentiable functions $f(x), g(x)$ satisfying $\displaystyle 6\int f(x)g(x)dx=x^{6}+3x^{4}+3x^{2}+c$ & $\displaystyle 2 \int \frac {g(x)dx}{f(x)}=x^{2}+c$. where $\displaystyle f(x)>0 \forall x \in R$