Mathematics

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

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

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

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Single Correct Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

#### 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. $\dfrac{3x^2}{2}+x +c$

1 Verified Answer | Published on 17th 09, 2020

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

1 Verified Answer | Published on 17th 09, 2020

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

1 Verified Answer | Published on 17th 09, 2020

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.

$\int {\left( {\frac{1}{{\left( {\ell nx} \right)}} - \frac{1}{{{{(\ell nx)}^2}}}} \right)\,dx}$