Question

integration of cos⁴x - sin⁴x/cosx - sinx × dx​

Answer 1

Answer:

Question:-

$\bf \:Evaluate: ∫\dfrac{ {cos}^{4} x - {sin}^{4} x}{cosx - sinx} dx$

Answer

Identity used :-

$\bf \: {x}^{4} - {y}^{4} = (x - y)(x + y)( {x}^{2} + {y}^{2} )$

$\bf \: {sin}^{2} x + {cos}^{2} x = 1$

$\bf \: ∫sinx \: dx = - cosx + c$

$\bf \: ∫cosx \: dx \: = sinx + c$

Solution:-

$\bf \:∫\dfrac{ {cos}^{4} x - {sin}^{4} x}{cosx - sinx} dx$

$\bf\implies \:∫\dfrac{(sinx + cosx)(cosx - sinx) ({sin}^{2} x + {cos}^{2} x)}{cosx - sinx} dx$

$\bf\implies \: ∫(sinx + cosx)dx$

$\bf\implies \: ∫sinx \: dx + ∫cosx \: dx$

$\bf\implies \: - cosx + sinx + c$

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