Mathematics

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


ANSWER

$$\cfrac { 1 }{ 32 } $$


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
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