Question

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Answer 1

$I = \frac{1}{2}( ln(cos(2x))-\frac{cos²(2x)}{2})+ C$

CALCULATION

$I = ∫ \frac{cos(4x)-1}{cot(x)-tan(x)}dx$

$I = - ∫ \frac{1-cos(4x)}{\frac{cos(x)}{sin(x)}-\frac{sin(x)}{cos(x)}}dx$

$I = -∫ \frac{2sin²(2x)}{\frac{cos²(x)-sin²(x)}{sin(x)cos(x)}}dx$

$I = -∫ \frac{sin²(2x).2sin(x)cos(x)}{cos(2x)}dx$

$I = -∫ \frac{1 - cos²(2x)}{cos(2x)}sin(2x)dx$

Substituting $cos(2x) = t$

$⇒ - 2sin(2x)dx = dt$

$I = ∫ \frac{1 - t²}{t}\frac{dt}{2}$

$I = \frac{1}{2}∫ (\frac{1}{t}-t)dt$

$I = \frac{1}{2}( ln(t)-\frac{t²}{2})+ C$

$I = \frac{1}{2}( ln(cos(2x))-\frac{cos²(2x)}{2})+ C$

Answer 2

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