Mathematics

# Integrate: $\int \dfrac{\sec^2\sqrt{x}}{\sqrt{x}}dx$

$I = 2 \tan \sqrt x+c$

##### SOLUTION
$I=\int { \frac { { sec }^{ 2 }\sqrt { x } }{ \sqrt { x } } } dx\\ Let\quad u=\sqrt { x } \Rightarrow du=\frac { 1 }{ 2\sqrt { x } } dx\\$
Substituting in I,

$I=2\int { { sec }^{ 2 }udu } \\ =2tanu+c\quad \quad \quad \left[ \because \frac { d\left( tanu \right) }{ du } { sec }^{ 2 }u \right]$

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

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

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Prove the following:
$\displaystyle \int [f(x)]^{n}f'(x) dx = \dfrac {[f(x)]^{n + 1}}{n + 1} + C$, if $n\neq -1$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Hard
Solve:
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Q3 Single Correct Hard
Evaluate $\displaystyle {\int \sin^{-1}\, \sqrt {\frac {x}{a\, +\, x}} dx}$
• A. $(a + x) \tan^{-1} \displaystyle \sqrt \frac {x}{a}\, -\, \sqrt {ax}\, +\, c$
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• C. $(a + x) \tan^{-1} \displaystyle \sqrt {x}\, -\, \sqrt {ax}\, +\, c$
• D. $\displaystyle \left(\frac { ax+1 }{ 2a } \right)\sin^{ -1 }\, \sqrt { \frac { x }{ a\, +\, x } } -\frac { 1 }{ 2 } \sqrt { \frac { x }{ a } } +c$

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Q4 Subjective Medium
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