Mathematics

# $Evalute\;\int {\dfrac{{\cot x}}{{\sqrt {\sin x} }}} dx.$

##### SOLUTION
Solution :-

$\displaystyle I = \int \frac{cotx}{\sqrt{sinx}}dx$

$\displaystyle = \int \frac{cosx}{sinx \sqrt{sinx}}dx$

Let $sinx = t$

$\therefore cosx\, dx = dt$
$\displaystyle \therefore I = \int \frac{dt}{t^{3/2}}$

$=- 2t^{-1/2}+C$

$= -\dfrac{2}{\sqrt{t}}+c$

$= \dfrac{-2}{\sqrt{sinx}}+c$

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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
Evaluate: $\displaystyle \int_{0}^{\pi}x\sin^{6}xdx$
• A. $\displaystyle \frac{35\pi^{2}}{1024}$
• B. $\displaystyle \frac{3\pi^{2}}{128}$
• C. $\displaystyle \frac{\pi^{2}}{32}$
• D. $\displaystyle \frac{5\pi^{2}}{32}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\displaystyle\int^{\pi/2}_{-\pi/2}\cos xdx=?$
• A. $0$
• B. $-1$
• C. None of these
• D. $2$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
If $\displaystyle I =\int _{\alpha}^{\beta} \left [\log \log x + \frac{1}{(\log x)^{2}} \right ] dx,$ then $I$
equals
• A. $\alpha \log \log \alpha -\beta \log \log \beta$
• B. $\displaystyle \frac{1}{\alpha}-\frac{1}{\beta}+\log \log \alpha -\log \log \beta$
• C. $\displaystyle \frac{\beta-\alpha}{\alpha \beta}+\alpha \log \log \alpha -\beta \log \log \beta$
• D. none of these

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
Solve: $\displaystyle \int^{\pi/2}_{0}(2log \sin x - log \sin 2x)dx$
• A. $-\frac{\pi}{2}log\frac{1}{2}$
• B. $\frac{\pi}{2}log{2}$
• C. $-{\pi}log\frac{1}{2}$
• D. $\frac{\pi}{2}log\frac{1}{2}$

Find $\int { x\sqrt { 1+x-{ x }^{ 2 } } dx }$