Mathematics

# If $\displaystyle\int { \cfrac { \cos { 8x } +1 }{ \tan { 2x } -\cot { 2x } } } dx=a\cos { 8x } +C\quad$, then $a=$

$\cfrac { 1 }{ 32 }$

##### SOLUTION
$I=\displaystyle\int{\dfrac{\cos{8x}+1}{\tan{2x}-\cot{2x}}dx}$

$=\displaystyle\int{\dfrac{2{\cos}^{2}{4x}-1+1}{\dfrac{\sin{2x}}{\cos{2x}}-\dfrac{\cos{2x}}{\sin{2x}}}dx}$

$=\displaystyle\int{\dfrac{2{\cos}^{2}{4x}}{\dfrac{{\sin}^{2}{2x}-{\cos}^{2}{2x}}{\sin{2x}\cos{2x}}}dx}$

$=\displaystyle\int{\dfrac{2{\cos}^{2}{4x}}{\dfrac{-2\cos{4x}}{\sin{2x}\cos{2x}}}dx}$

$=\displaystyle\int{\dfrac{{\cos}^{2}{4x}\sin{4x}}{-2\cos{4x}}dx}$

$=\displaystyle\int{\dfrac{2\cos{4x}\sin{4x}}{-4}dx}$

$=\displaystyle\int{\dfrac{\sin{8x}}{-4}dx}$

$=-\dfrac{1}{4}\displaystyle\int{\sin{8x}dx}$

$=-\dfrac{1}{4}\left[\dfrac{-\cos{8x}}{8}\right]+c$

$=\dfrac{1}{32}\cos{8x}+c$

Comparing with $a\cos{8x}+c$ we get$\ a=\dfrac{1}{32}$

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Single Correct Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 114

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