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If $\int \frac{2\cos x-\sin x+\lambda }{\cos x+\sin x-2}dx=A\: ln\left | \cos x+\sin x-2 \right |+Bx+C$ Then the ordered triplet $A,B,\lambda$ is
Option: 1
$\left ( \frac{1}{1},\frac{3}{2},-1 \right )$

Option: 2
$\left ( \frac{3}{2},\frac{1}{2},-1 \right )$

Option: 3
$\left ( \frac{1}{1},-1,-\frac{3}{2} \right )$

Option: 4
$\left ( \frac{3}{2},-1,\frac{1}{2} \right )$

Answers (1)

best_answer

Type of Integration -

Integrals in the form of $\int \frac{p\cos x+q\sin x }{a\cos x+b\sin x}dx$

- wherein

Working rule :

$p\cos x+q\sin x=A\cdot \frac{\mathrm{d} }{\mathrm{d} x}(a\cos x+b\sin x)+B(a\cos x+b\sin x)$

Find $A$ and $B$ by comparing $sinx$ and $cosx$

$\frac{d}{dx}(A\: ln\left | \cos x+\sin x-2 \right |+Bx+C)$

$=A\frac{\cos x-\sin x}{\cos x+\sin x-2}+B$

$=\frac{A\cos x-A\sin x+Bcos x+B\sin x-2B}{\cos x+\sin x-2}$

$\therefore 2=A+B,-1=A+B,\lambda =-2B$

$\therefore A=3/2,B=1/2,\lambda =-1$

Posted by

Suraj Bhandari

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