Mathematics

# Evaluate the following integrals$\int { \cfrac { 1 }{ 2+\sin { x } +\cos { x } } } dx$

##### SOLUTION
Let $\text{I}=\displaystyle\int \dfrac{1}{2+\sin x+\cos x} d x$

$=\displaystyle\int \dfrac{1}{2+\bigg(\dfrac{2\tan (x/2)}{1+\tan^2 (x/2)}\bigg)+\bigg(\dfrac{1-\tan^2 x/2}{1+\tan^2 x/2}\bigg)}d x$       $\bigg(\because \sin 2 A=\dfrac{2\tan A}{1+\tan^2 A},\cos 2 A=\dfrac{1-\tan^2 A}{1+\tan^2 A}\bigg)$

$=\displaystyle\int \dfrac{1+\tan^2 \dfrac{x}{2}}{2\bigg(1+\tan^2 \dfrac{x}{2}\bigg)+\bigg(1-\tan^2 \dfrac{x}{2}\bigg)+2\tan \dfrac{x}{2}}d x$

$=\displaystyle\int \dfrac{\text{sec}^2 \dfrac{x}{2}}{3+2\tan \dfrac{x}{2}+\tan^2 \dfrac{x}{2}}d x$               $(\because \text{sec}^2 A-\tan^2 A=1)$

Let $t=\tan \dfrac{x}{2}\implies d t=\dfrac{1}{2}\text{sec}^2 \dfrac{x}{2} d x$

So $I=\displaystyle\int \dfrac{2 d t}{ t^2+2 t+3}$

$=2\displaystyle\int \dfrac{d t}{t^2+2(t)(1)+3}$

$=2\displaystyle\int \dfrac{d t}{t^2+2(t)(1)+(1)^2+3-(1)^2}$

$=2\displaystyle\int \dfrac{d t}{(t+1)^2-2}$                    $(\because (a+b)^2=a^2+2 a b+b^2\ )$

$=2\displaystyle\int \dfrac{d t}{(t+1)^2+(\sqrt{2})^2}$

$=\dfrac{2}{\sqrt{2}}\text{tan}^{-1} \dfrac{t+1}{\sqrt{2}}+c$                                        $\bigg(\because \displaystyle\int \dfrac{d x}{x^2+a^2}=\dfrac{1}{a}\text{tan}^{-1} \dfrac{x}{a}+c\ \bigg)$

$=\sqrt{2} \text{tan}^{-1} \bigg(\dfrac{1}{\sqrt{2}}\tan \dfrac{x}{2}+\dfrac{1}{\sqrt{2}}\bigg)+c$                       $\bigg(\because t=\tan \dfrac{x}{2}\bigg)$

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

#### Realted Questions

Q1 Single Correct Medium
$\displaystyle \int \frac{\sec x}{\log \left ( \sec x+\tan x \right )}dx$
• A. $\displaystyle \log \left [ \log \left ( \sec x-\tan x \right ) \right ].$
• B. $\displaystyle \log \left [ \log \left ( \sin x+\cos x \right ) \right ].$
• C. $\displaystyle \log \left [ \log \left ( \sin x-\cos x \right ) \right ].$
• D. $\displaystyle \log \left [ \log \left ( \sec x+\tan x \right ) \right ].$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
$\displaystyle \int e^{x}[\log(x+\sqrt{x^{2}-1})+\frac{d}{dx}(\cos h^{-1}x)]dx=$
• A. $e^{x}\sin h^{-1}x+c$
• B. $e^{x} \log(x+\sqrt{x^{2}+1})+c$
• C. $-e^{x}\sinh^{-1}x+c$
• D. $e^{x}\cosh^{-1}x+c$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
If $\displaystyle \int \dfrac{e^x - 1}{e^x + 1}dx =f(x) + c$ then $f(x) =$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
If $\displaystyle I = \int \frac {\cos x \: dx}{\sqrt {a + b \cot^2 x}} (a > b > 0)$, then I equals
• A. $\displaystyle \frac {1}{a - b} \sqrt {a + b \cot^2 x} + C$
• B. $\displaystyle \frac {1}{a - b} \left ( \sqrt {a + b \cot^2 x} + x \right ) + C$
• C. $\displaystyle \frac {1}{a - b} \left ( \sqrt {a + b \cot^2 x} - x \right ) + C$
• D. $\displaystyle \frac {1}{a - b} \sqrt {a \sin^2 x + b \cos^2 x} + C$

1 Verified Answer | Published on 17th 09, 2020

Q5 Single Correct Hard
$\lim_{n\rightarrow \infty } \left ( \dfrac{n}{n^{2}}+\dfrac{n}{n^{2}+1}+\dfrac{n}{n^{2}+2^{2}}+......+\dfrac{n}{2n^{2}-2n+1} \right )$ is equal to
• A. 1
• B. tan 1
• C. $tan \dfrac{\pi }{4}$
• D. $\dfrac{\pi }{4}$