Mathematics

# If $\int\limits_0^1 {\frac{{\sin t}}{{1 + t}}} dt = \alpha ,\,then\,\int\limits_{4\pi - 2}^{4\pi } {\frac{{\sin (t/2)}}{{4\pi + 2 - t}}dt}$ in terms of $\alpha$ is given by

$- \alpha$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

#### Realted Questions

Q1 Single Correct Medium
$\left[ \int \log ( 1 + \cos x ) - x \tan \frac { x } { 2 } \right] d x$ is equal to ?
• A. $x \tan \frac { x } { 2 }$
• B. $x \log ( 1 + \cos x )$
• C. None of these
• D. $\log ( 1 + \cos x )$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Evaluate :
$\int \dfrac{(\log x)^3}{x} \ dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Hard
$\int {{{\tan }^{ - 1}}\left( {\sqrt {\frac{{1 - x}}{{1 + x}}} } \right)dx}$ equals?

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$\displaystyle \frac{2x^{2}+2x+1}{x^{3}+x^{2}}=$
• A. $\displaystyle \dfrac{1}{x}-\dfrac{1}{x^{2}}-\dfrac{1}{x+1}$
• B. $\dfrac{1}{x}-\dfrac{1}{x^{2}}+\displaystyle \frac{1}{x+1}$
• C. $\displaystyle \frac{1}{x}+\frac{1}{x^{2}}-\frac{1}{x+1}$
• D. $\dfrac{1}{x}+\dfrac{1}{x^{2}}+\displaystyle \frac{1}{x+1}$

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 $\sqrt{mx^2+nx+p}=(x-\alpha)u\:or\:(x-\beta)u$