Mathematics

# Solve $\displaystyle\int\dfrac {\sqrt {\tan x}}{\sin x \cos x}dx$

##### SOLUTION
$\displaystyle\int\dfrac {\sqrt {\tan x}}{\sin x \cos x}dx$
Put, $\tan x = t$

$\implies \sec^2 x dx = dt$

Now, $\displaystyle\int \dfrac{\sqrt{\tan x}}{\sin x \cos x} dx = \int \dfrac{\sqrt{\tan x}}{\tan x \cos^2 x} dx$

$= \displaystyle\int \dfrac{\sec^2 x}{\sqrt{\tan x}} dx = \int \dfrac{1}{\sqrt{t}} dt$

$= 2\sqrt{t} + c$

$= 2\sqrt{\tan x} + c$

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

#### Realted Questions

Q1 Single Correct Medium
$\displaystyle \int e^{x}(cotx-cot^{2}x)dx=$
• A. $(\cot x) +c$
• B. $e^{x}(1-\cot^{2}x)+c$
• C. $e^{x}\cot^{2}x+c$
• D. $e^{x}(\cot x+1)+c$

1 Verified Answer | Published on 17th 09, 2020

Q2 Multiple Correct Medium
Let $\displaystyle S_{n}=\frac{n}{\left ( n+1 \right )\left ( n+2 \right )}+\frac{n}{\left ( n+2 \right )\left ( n+4 \right )}+\frac{n}{\left ( n+3 \right )\left ( n+6 \right )}+......+\frac{1}{6n}$, then $\displaystyle \lim_{n\rightarrow \infty }S_{n}$ is
• A. $\displaystyle \ln\frac{9}{2}$
• B. $>1$
• C. $\displaystyle \ln\frac{3}{2}$
• D. $<2$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
$\displaystyle \int \frac{\tan 2\theta d\theta }{\sqrt{\cos ^{6}\theta +\sin ^{6}\theta }}$ is equal to
• A. $\ln\left |\displaystyle \frac{\sqrt{1+3\cos ^{2}2\theta }}{\cos 2\theta } \right |+c$
• B. $\ln\left |\displaystyle \frac{1+\sqrt{2+3\cos ^{2}2\theta }}{\sin 2\theta } \right |+c$
• C. None of these
• D. $\ln\left |\displaystyle \frac{1+\sqrt{1+3\cos ^{2}2\theta }}{\cos 2\theta } \right |+c$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
Taking constant of integration as zero, find f(1).
$\displaystyle \int\frac{xe^{x}}{\left ( x+1 \right )^{2}}dx$

Let $\displaystyle f\left ( x \right )=\frac{\sin 2x \cdot \sin \left ( \dfrac{\pi }{2}\cos x \right )}{2x-\pi }$