Mathematics

# $\displaystyle \int \frac{\sec x\:co\sec x }{\log \tan x}dx$

$\displaystyle\log \left ( \log \tan x \right ).$

##### SOLUTION
Let $\displaystyle I=\int \frac{\sec x\:cosec x }{\log \tan x}dx$
Put $\displaystyle \log \tan x=t\Rightarrow \sec xcosec xdx=dt$
Therefore
$\displaystyle I=\int \frac { dt }{ t } =\log t=\log \left( \log \tan x \right)$
Hence, option 'C' is correct.

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

#### Realted Questions

Q1 Subjective Hard
Find $\displaystyle \int \dfrac {1}{\sin x\cos^{3}x}dx.$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
Solve: $\displaystyle\int \dfrac{\sin 2x}{\sin^4 x+\cos^4 x} dx=$
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• C. $\tan^{-1} (2\tan x)$
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Q3 Subjective Medium
Evaluate: $\displaystyle \int _0^1 (8x^2+16) dx$

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Q4 Single Correct Medium
$\int _{ -1 }^{ 1/2 }{ \dfrac { { e }^{ x }\left( 2-{ x }^{ 2 } \right) dx }{ \left( 1-x \right) \sqrt { 1-{ x }^{ 2 } } } }$ is equal to
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Q5 Single Correct Medium
The value of the integers $\int\limits_0^2 {\dfrac{{\log ({x^2} + 2)}}{{{{\left( {x + 2} \right)}^2}}}dx\,is:}$
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