Question

$\int \frac{\cos x+3\sin x+7}{\cos x+\sin x+1}dx$ is equal to

• A. $-\log |\cos x+\sin x+1|+2x+5\log |1+\tan \frac{x}{2}|+c$
• B. $\log |\cos x+\sin x+1|+2x+5\log |1+\tan \frac{x}{2}|+c$
• C. $\log |\cos x+\sin x+1|+2x-5\log |1+\tan \frac{x}{2}|+c$
• D. $-\log |\cos x+{\sin }_X+1|+2x-5\log |1+\tan \frac{x}{2}|+c$

Answer 1

The correct option is A $-\log |\cos x+\sin x+1|+2x+5\log |1+\tan \frac{x}{2}|+c$

$I=\int \frac{\cos x+3\sin x+7}{\cos x+\sin x+1}dx$
We can write
$\cos x+3\sin x+7=A(\cos x+\sin x+1)+B\frac{d}{dx}(\cos x+\sin x+1)+C$
$\cos x+3\sin x+7=A(\cos x+\sin x+1)+B(-\sin x+\cos x)+C$ ....(1)
$\cos x+3\sin x+7=(A+B)\cos x+(A-B)\sin x+(A+C)$

On comparing,
$A+B=1,A-B=3,A+C=7$

Solving these equations, we get
$A=2,B=-1,C=5$

Put these values in (1)
$\cos x+3\sin x+7=2(\cos x+\sin x+1)-1(-\sin x+\cos x)+5$

So, $\int \frac{\cos x+3\sin x+7}{\cos x+\sin x+1}dx=2\int \frac{\cos x+\sin x+1}{\cos x+\sin x+1}dx-\int \frac{(-\sin x+\cos x)}{\cos x+\sin x+1}dx+5\int \frac{dx}{\cos x+\sin x+1}$
$\int \frac{\cos x+3\sin x+7}{\cos x+\sin x+1}dx=2\int dx-\int \frac{(-\sin x+\cos x)}{\cos x+\sin x+1}dx+5\int \frac{dx}{\cos x+\sin x+1}$
Put$\cos x+\sin x+1=t$
$\Rightarrow (-\sin x+\cos x)dx=dt$
$\int \frac{\cos x+3\sin x+7}{\cos x+\sin x+1}dx=2x-\int \frac{1}{t}dt+5\int \frac{dx}{\frac{1-{\tan }^{2}x/2}{1+{\tan }^{2}x/2}+\frac{2\tan x/2}{1+{\tan }^{2}x/2}+1}$
$\int \frac{\cos x+3\sin x+7}{\cos x+\sin x+1}dx=2x-\log |t|+5\int \frac{{\sec }^{2}x/2dx}{2(1+\tan x/2)}$
Put$(1+\tan x/2)=u$
$\frac{1}{2}{\sec }^{2}x/2dx=du$
$\int \frac{\cos x+3\sin x+7}{\cos x+\sin x+1}dx=2x-\log |(\cos x+\sin x+1)|+5\int \frac{du}{u}$
$\int \frac{\cos x+3\sin x+7}{\cos x+\sin x+1}dx=2x-\log |(\cos x+\sin x+1)|+5\log u+c$

$=-\log |\cos x+\sin x+1|+2x+5\log |1+\tan \frac{x}{2}|+c$