Mathematics

$$\int \displaystyle \frac {cos^2\, x}{1+tan\, x}\, dx$$


ANSWER

$$\displaystyle \frac {1}{4}\, ln\, (cos\, +\, sin\, x)\, +\, \displaystyle \frac {x}{2}\, +\, \displaystyle \frac {1}{8}\, (sin\, 2x\, +\, cos\, 2x)$$


SOLUTION
Let $$\displaystyle I=\int { \frac { \cos ^{ 2 }{ x }  }{ 1+\tan { x }  } dx } $$
Multiply numerator and denominator by $$\sec ^{ 4 }{ x } $$

$$\displaystyle I=\int { \frac { \sec ^{ 2 }{ x }  }{ \left( \tan { x } +1 \right) { \left( \tan ^{ 2 }{ x } +1 \right)  }^{ 2 } } dx } $$
Put $$t=\tan { x } \Rightarrow dt=\sec ^{ 2 }{ x } dx$$

$$\displaystyle I=\int { \frac { 1 }{ \left( t+1 \right) { \left( { t }^{ 2 }+1 \right)  }^{ 2 } } dt } =\int { \left( \frac { 1-t }{ 4\left( { t }^{ 2 }+1 \right)  } +\frac { 1-t }{ 2{ \left( { t }^{ 2 }+1 \right)  }^{ 2 } } +\frac { 1 }{ 4\left( t+1 \right)  }  \right) dt } $$

$$\displaystyle =-\frac { 1 }{ 4 } \int { \frac { t }{ { t }^{ 2 }+1 } dt } +\frac { 1 }{ 4 } \int { \frac { 1 }{ { t }^{ 2 }+1 } dt } +\frac { 1 }{ 2 } \int { \frac { 1-t }{ { \left( { t }^{ 2 }+1 \right)  }^{ 2 } } dt } +\frac { 1 }{ 4 } \int { \frac { 1 }{ t+1 } dt } $$

$$\displaystyle =-\frac { 1 }{ 8 } \log { \left( { t }^{ 2 }+1 \right)  } +\frac { 1 }{ 4 } \tan ^{ -1 }{ \left( { t }^{ 2 }+1 \right)  } +\frac { 1 }{ 4 } \log { \left( t+1 \right)  } +{ I }_{ 1 }$$
Where $$\displaystyle { I }_{ 1 }=\frac { 1 }{ 2 } \int { \frac { 1-t }{ { \left( { t }^{ 2 }+1 \right)  }^{ 2 } } dt } $$

Put $$t=\tan { u } \Rightarrow dt=\sec ^{ 2 }{ u } du$$
$$\displaystyle { I }_{ 1 }=\frac { 1 }{ 2 } \int { \cos ^{ 2 }{ u } \left( 1-\tan { u }  \right) du } =\frac { 1 }{ 2 } \int { \left( 1-\sin ^{ 2 }{ u }  \right) \left( 1-\tan { u }  \right) du } $$

$$\displaystyle =\frac { 1 }{ 2 } \int { \left( -\sin ^{ 2 }{ u } -\tan { u } +\sin ^{ 2 }{ u } \tan { u } +1 \right) du } =\frac { u }{ 4 } +\frac { 1 }{ 8 } \sin { 2u } +\frac { \cos ^{ 2 }{ u }  }{ 4 } $$

Therefore 
$$\displaystyle I=\frac { 1 }{ 8 } \left( \sin { 2x } +\cos { 2x } +4\tan ^{ -1 }{ \tan { x }  } +2\log { \left( \tan { x } +1 \right)  } -\log { \left( \sec ^{ 2 }{ x }  \right)  } +1 \right) $$
$$\displaystyle =\frac { x }{ 2 } +\frac { 1 }{ 8 } \left( \sin { 2x } +\cos { 2x }  \right) +\frac { 1 }{ 4 } \log { \left( \sin { x } +\cos { x }  \right)  } $$

View Full Answer

Its FREE, you're just one step away


Single Correct Hard Published on 17th 09, 2020
Next Question
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105
Enroll Now For FREE

Realted Questions

Q1 Single Correct Medium
Integrate w.r.t to $$x$$
$$ {3x+1}$$
  • A. $$x^3+3x+c$$
  • B. $$3x+c$$
  • C. None of these
  • D. $$\dfrac{3x^2}{2}+x +c$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q2 Subjective Medium
Solve $$xdy-ydx=x^3dy+x^2ydx$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q3 TRUE/FALSE Hard
$$\displaystyle \int_{-2}^3 (x +1)dx=\displaystyle \lim_{n \rightarrow \infty} \displaystyle \Sigma_1^n \left(\dfrac{5i}{n}- 1 \right)$$

State whether the above statment is True or False?
  • A. True
  • B. False

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q4 Single Correct Hard
$$\displaystyle \int_{0}^{4}\frac{(y^2-4y+5)\sin (y-2)dy}{[2y^2-8y+11]}$$ is equal to
  • A. 2
  • B. -2
  • C. none of these
  • D.

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q5 Subjective Medium
$$\int {\left( {\frac{1}{{\left( {\ell nx} \right)}} - \frac{1}{{{{(\ell nx)}^2}}}} \right)\,dx} $$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer