Mathematics

# Evaluate$\displaystyle\int {\dfrac{{\sin x + \cos x}}{{{{\left( {\sin x - \cos x} \right)}^2}}}dx}$

$\dfrac {-1}{\sin x-\cos x}+C$

##### SOLUTION
$\int \dfrac{\sin x + \cos x}{(\sin x - \cos x)^2} dx$
Let $\sin x - \cos x = t$
$\dfrac{d}{dx} (\sin x - \cos x) = \dfrac{dt}{dx}$
$(\cos x + \sin x)dx = dt$
$\int \dfrac{dt}{t^2}$
$\int t^{-2} dt$
$= -t^{-1} + C = \dfrac{-1}{\sin x - \cos x} + C$
$\dfrac{d}{dx} (\sin x) = \cos x$
$\dfrac{d}{dx} (\cos x) = \sin x$
$\int x^n dx = \dfrac{x^{n+1}}{n+1}+c$

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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 Medium
$\displaystyle\int\dfrac{e^x}{x}(1+x\log |x|)\ dx=$
• A. $e^x\log|x|+e^x+c$
• B. $e^x\log|s|-x+c$
• C. none of these
• D. $e^x\log |x|+c$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
If $\displaystyle I = \int \frac {dx}{x^4 \sqrt {1 + x^2}}$, then I equals
• A. $\displaystyle \frac {\sqrt {x^2 + 1}}{x} - \frac {1}{2x^2} + C$
• B. $\displaystyle \frac {\sqrt {1 + x^2}}{x} - \frac {1}{2x^3} + C$
• C. $\displaystyle - \frac {\sqrt {1 + x^2}}{x} + \frac {2x}{\sqrt {1 + x^2}} + C$
• D. none of these

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Hard
Integrate:
$\int { \frac { dx }{ 3cosx+4sinx+6 } }$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
The value of $\displaystyle \int { { e }^{ 2x } } \left( \frac { 1 }{ x } -\frac { 1 }{ 2{ x }^{ 2 } } \right) dx$ is
• A. $\displaystyle \frac { { e }^{ 2x } }{ 2 } +c$
• B. $\displaystyle \frac { { e }^{ 2x } }{ 3x } +c$
• C. $\displaystyle \frac { { e }^{ 2x } }{ x } +c$
• D. $\displaystyle \frac { { e }^{ 2x } }{ 2x } +c$

1 Verified Answer | Published on 17th 09, 2020

Q5 Passage Hard
Let us consider the integral of the following forms
$f{(x_1,\sqrt{mx^2+nx+p})}^{\tfrac{1}{2}}$
Case I If $m>0$, then put $\sqrt{mx^2+nx+C}=u\pm x\sqrt{m}$
Case II If $p>0$, then put $\sqrt{mx^2+nx+C}=u\pm \sqrt{p}$
Case III If quadratic equation $mx^2+nx+p=0$ has real roots $\alpha$ and $\beta$, then put $\sqrt{mx^2+nx+p}=(x-\alpha)u\:or\:(x-\beta)u$