Mathematics

# Evaluate: $\displaystyle\int {\dfrac{1+\cos 4x}{\cot x-\tan x}dx}$

##### SOLUTION
$\int { \cfrac { 1+\cos { 4x } }{ \cot { x } -\tan { x } } } dx=\int { \cfrac { 1+2\cos ^{ 2 }{ 2x } -1 }{ \cfrac { \cos { x } }{ \sin { x } } -\cfrac { \sin { x } }{ \cos { x } } } } dx$
$=\int { \cfrac { 2\cos ^{ 2 }{ 2x } .\sin { x } \cos { x } }{ \cos ^{ 2 }{ x }- \sin ^{ 2 }{ x } } } dx=\int { \cfrac { \cos ^{ 2 }{ 2x } .\sin { 2x } }{ \cos { 2x } } } dx$
$=\int { \cos { 2x } .\sin { 2x } } dx=\cfrac { 1 }{ 2 } \int { 2\sin { 2x } .\cos { 2x } } dx$
$=\cfrac { 1 }{ 2 } \int { \sin { 4x } } dx=-\cfrac { 1 }{ 8 } \cos { 4x } +C$

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

#### Realted Questions

Q1 Single Correct Medium

$\displaystyle \int_{1}^{\infty}\left( \frac { 1 }{ 1+x^{ 2 } } \right) d{ x }=$

• A. $-\displaystyle \frac{\pi}{4}$
• B. $\displaystyle \frac{\pi}{2}$
• C. $-\displaystyle \frac{\pi}{2}$
• D. $\displaystyle \frac{\pi}{4}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Evaluate $\displaystyle\int^{\pi/6}_0\sec^2xdx$.

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
The value of $\displaystyle \lim_{n\to\infty} \displaystyle \sum_{r=1}^{n}\frac{1}{n}\sqrt{\frac{n+r}{n-r}}$ is
• A. $\displaystyle \frac{\pi}{2}$
• B. $2\pi$
• C. $\displaystyle \frac{\pi}{2}-1$
• D. $\displaystyle \frac{\pi}{2}+1$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Solve
$\int (sinx + cos x )^2. dx$

1 Verified Answer | Published on 17th 09, 2020

Q5 Single Correct Medium
If $\dfrac{3x^{2}+10x+13}{(x-1)^{4}}=\dfrac{A}{(x-1)^{2}}+\dfrac{B}{(x-1)^{3}}+\dfrac{C}{(x-1)^{4}}$ then descending order of $A,B,C$
• A. $A, B, C$
• B. $A, C, B$
• C. $C, A, B$
• D. $C, B, A$