Mathematics - Indefinite Integration Question with Solution | TestHub

To refine the solution, we differentiate the given integral result and equate the coefficients of $\cos x$, $\sin x$, and the constant term in the numerator.

Given:

$$\int \frac{2\cos x-\sin x+\lambda }{\cos x+\sin x-2}dx=A\ln |\cos x+\sin x-2|+Bx+C$$

Differentiating both sides with respect to $x$:

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

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

Equating the numerators:

$$2\cos x-\sin x+\lambda =(A+B)\cos x+(-A+B)\sin x-2B$$

Comparing coefficients:

For $\cos x$: $A+B=2$

For $\sin x$: $-A+B=-1$

For constant term: $-2B=\lambda$

Adding the first two equations: $(A+B)+(-A+B)=2+(-1)⟹2B=1⟹B=\frac{1}{2}$

Substituting $B=\frac{1}{2}$ into $A+B=2$: $A+\frac{1}{2}=2⟹A=\frac{3}{2}$

Substituting $B=\frac{1}{2}$ into $-2B=\lambda$: $\lambda =-2\left(\frac{1}{2}\right)⟹\lambda =-1$

Thus, the ordered triplet $(A,B,\lambda )$ is $\left(\frac{3}{2},\frac{1}{2},-1\right)$.