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Question

Solve $\int \frac{cos 4 x - cos 4 x}{sin 4 x - sin 2 x} d x$

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Solution

According to question,

$\int \frac{c o s 4 x - c o s 4 x}{s i n 4 x - s i n 2 x} d x$

$⟹ \int \frac{0}{s i n 4 x - s i n 2 x} d x$

$⟹ \int 0. d x$

$⟹ 0$

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Similar questions

\int \frac{{cos}^{4} x - {sin}^{4} x}{\sqrt{1 + sin 2 x}} d x

\int \frac{cos 4 x - cos 2 x}{sin 4 x - sin 2 x} d x

{cos}^{4} x - {sin}^{4} x =

\int (\frac{tan x d x}{\sqrt{{sin}^{4} x + {cos}^{4} x}})

{sin}^{4} x + {cos}^{4} x = 5 / 8

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