Question

Integrate cosx-2sinx/sinx + cosx dx​

Answer 1

Answer:

$\tt{ \int {\dfrac{cosx-2sinx}{sinx+cosx} } \, dx = \dfrac{-x}{2} + \dfrac{3}{2} \, ln|sin\,x+cos\,x| + c }$

Step-by-step explanation:

To find: $\tt{ \int {\dfrac{cosx-2sinx}{sinx+cosx} } \, dx }$

Solution:

Let I = $\tt{ \int {\dfrac{cosx-2sinx}{sinx+cosx} } \, dx }$

w.k.t., to solve integrals of type $\tt{ \int {\dfrac{p \, cosx+q \, sinx}{a \, sinx+b \, cosx} } \, dx }$, we substitute numerator according to following formula :-

Numerator = A×(Denominator) + B×(derivative of Denominator)

⇒ $\tt{ cos\,x - 2sin\,x = A\times(sin\,x+cos\,x) + B\times\dfrac{d}{dx}(sin\,x+cos\,x)}$

⇒ $\tt{ cos\,x - 2sin\,x = A\,sin\,x+A\,cos\,x + B(cos\,x-sin\,x)}$

⇒ $\tt{ cos\,x - 2sin\,x = A\,sin\,x+A\,cos\,x + B\,cos\,x-B\,sin\,x}$

⇒ $\tt{ cos\,x - 2sin\,x = (A-B)\,sin\,x+(A+B)\,cos\,x }$

On compairing both sides, we get

$\tt{ A-B = -2 \,\, and \,\, A+B = 1 }$

On solving above two equations, we get A = -1/2 and B = 3/2.

$\tt{\implies I = \int {\dfrac{A(\,sin\,x+\,cos\,x) + B(cos\,x-sin\,x)}{sinx+cosx} } \, dx }$

$\tt{\implies I = \int { [\dfrac{A(\,sin\,x+\,cos\,x)}{sinx+cosx} + \dfrac{B(cos\,x-sin\,x)}{sinx+cosx} } ] \, dx }$

$\tt{\implies I = \int { [A + \dfrac{B(cos\,x-sin\,x)}{sinx+cosx} } ] \, dx }$

$\tt{\implies I = \int { [\dfrac{-1}{2} + \dfrac{\dfrac{3}{2}(cos\,x-sin\,x)}{sinx+cosx} } ] \, dx }$

$\tt{\implies I = \int { \dfrac{-1}{2} \, dx + \int{ \, \dfrac{3(cos\,x-sin\,x)}{2(sinx+cosx)} } \, dx }$

$\tt{\implies I = \int { \dfrac{-1}{2} \, dx + \dfrac{3}{2} \int{ \, \dfrac{(cos\,x-sin\,x)}{(sinx+cosx)} } \, dx }$

$\tt{\implies I = \dfrac{-x}{2} + \dfrac{3}{2} \, ln|sin\,x+cos\,x| + c }$

Formulas used :-

1) $\tt{ \int{ \dfrac{f^{'}(x)}{f(x)} } = ln|f(x)| + c }$