Mathematics

# $\displaystyle \int \frac{1}{\cos ^{2}x\left ( 1-\tan x \right )^{2}}dx.$

$\displaystyle \frac{1}{1-\tan x}.$

##### SOLUTION
Let $\displaystyle I=\int \frac { 1 }{ \cos ^{ 2 } x\left( 1-\tan x \right) ^{ 2 } } dx=\int \frac { \sec ^{ 2 } xdx }{ \left( 1-\tan x \right) ^{ 2 } }$

Put $\displaystyle 1-\tan x=t\Rightarrow -\sec ^{ 2 } xdx=dt$
Therefore
$\displaystyle I=-\int \frac { 1 }{ t^{ 2 } } dt=-\left( -\frac { 1 }{ t } \right) =\frac { 1 }{ t } =\frac { 1 }{ 1-\tan x }$
Hence, option 'A' 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 Single Correct Medium
$\int _{ 0 }^{ \pi }{ x\ln { \left( \sin { x } \right) } } dx=$
• A. $\dfrac { \pi }{ 2 } \ln { 2 }$
• B. $\dfrac { -\pi }{ 2 } \ln { 2 }$
• C. $-2p\ \ln { 2 }$
• D. $\dfrac { -\pi^{2} }{ 2 } \ln { 2 }$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Evaluate the following integral:
$\displaystyle\int^1_0\dfrac{2x}{(1+x^4)}dx$.

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
$\int { \cfrac { 1 }{ 8\sin ^{ 2 }{ x } +1 } } dx$ is equal to
• A. $\sin ^{ -1 }{ \left( \tan { x } \right) } +C$
• B. $\cfrac { 1 }{ 3 } \sin ^{ -1 }{ \left( \tan { x } \right) } +C$
• C. $\tan ^{ -1 }{ \left( 3\tan { x } \right) } +C$
• D. $\sin ^{ -1 }{ \left( 3\tan { x } \right) } +C$
• E. $\cfrac { 1 }{ 3 } \tan ^{ -1 }{ \left( 3\tan { x } \right) } +C$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
If $\displaystyle I = \int \frac {\cos x}{\sin^3 x - \cos^3 x} dx$, and $\displaystyle f(x) = \frac {1}{3} \log (\tan x - 1) + \frac {1}{\sqrt 3} \tan^{-1} \left ( \frac {2 \tan x + 1}{\sqrt 3} \right ) g(x) = \frac {1}{6} \log (\tan^2 x + \tan x + 1)$ then I equals
• A. $\displaystyle f(x) g(x) + C$
• B. $\displaystyle f(x)/g(x) + C$
• C. $\displaystyle f(x) + g(x) + C$
• D. $\displaystyle f(x) - g(x) + C$

Evaluate $\displaystyle \int [ \sqrt { \cot x } + \sqrt { \tan x } ] d x$