Question

evauluate the integral of int (cos⁴x-sin⁴x/cosx-sinx)dx
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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:-

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

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

$\bf\implies \: ∫(sinx + cosx)dx$

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

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

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