Mathematics

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


ANSWER

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


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.
View Full Answer

Its FREE, you're just one step away


Single Correct Medium Published on 17th 09, 2020
Next Question
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84
Enroll Now For FREE

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. $$\dfrac { -\pi^{2} }{ 2 } \ln { 2 } $$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

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

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
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. $$\cfrac { 1 }{ 3 } \tan ^{ -1 }{ \left( 3\tan { x } \right) } +C$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
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. $$\displaystyle f(x) - g(x) + C$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q5 Subjective Hard
Evaluate $$\displaystyle \int [ \sqrt { \cot x } + \sqrt { \tan x } ] d x$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer