Question

$I:\int \frac{a\cos x+b\sin x}{c\cos x+d\sin x}dx=$$\frac{ac+bd}{c^2+d^2}x+\frac{ad-bc}{c^2+d^2}\log |c\cos x+d\sin x|+k$
$II:\int \frac{\cos x+\sin x}{\sin x-\cos x}dx=\log |\sin x-\cos x|+k$ Which of the following is true

• A. Only I
• B. Only II
• C. Both I and II
• D. neither I nor II are true

Answer 1

The correct option is C Both I and II

Statement I $\int \frac{a\cos x+b\sin x}{c\cos x+d\sin x}dx=$$\frac{ac+bd}{c^2+d^2}x+\frac{ad-bc}{c^2+d^2}\log |c\cos x+d\sin x|+k$
Let $I=\int \frac{a\cos x+b\sin x}{c\cos x+d\sin x}dx$
Here numerator can be written as
$a\cos x+b\sin x=A(c\cos x+d\sin x)+B(-c\sin x+d\cos x)$ ....(1)
$\Rightarrow a\cos x+b\sin x=(Ac+Bd)\cos x+(Ad-Bc)\sin x$
On comparing , we get
$Ac+Bd=a$
$Ad-Bc=b$
Solving these equations, we get
$A=\frac{ac+bd}{d^2+c^2}$ ; $B=\frac{ad-bc}{d^2+c^2}$
Substituting these values in (1), we get
$a\cos x+b\sin x=\frac{(ac+bd)}{d^2+c^2}(c\cos x+d\sin x)+\frac{(ad-bc)}{d^2+c^2}(-c\sin x+d\cos x)$
So, the given integral becomes
$I=\frac{(ac+bd)}{d^2+c^2}\int 1dx+\frac{(ad-bc)}{d^2+c^2}\int \frac{(-c\sin x+d\cos x)}{c\cos x+d\sin x}dx$
Put $c\cos x+d\sin x=t$
$(-c\sin x+d\cos x)dx=dt$
$I=\frac{(ac+bd)}{d^2+c^2}x+\frac{(ad-bc)}{d^2+c^2}\int \frac{1}{t}dt$
$I=\frac{(ac+bd)}{d^2+c^2}x+\frac{(ad-bc)}{d^2+c^2}\log |c\cos x+d\sin x|+k$
Statement II: $\int \frac{\cos x+\sin x}{\sin x-\cos x}dx=\log |\sin x-\cos x|+k$
Let $I=\int \frac{\cos x+\sin x}{\sin x-\cos x}dx$
Put $\sin x-\cos x=t$
$\Rightarrow (\cos x+\sin x)dx=dt$
$I=\int \frac{1}{t}dt$
$I=\log |t|+k$
$\Rightarrow I=\log |\sin x-\cos x|+k$
Hence, both statements are correct