Question
Evaluvate the following intregals:
$\int\frac{3+2\cos\text{x}+4\sin\text{x}}{2\sin\text{x}+\cos\text{x}+3}\ \text{dx}$
$\int\frac{3+2\cos\text{x}+4\sin\text{x}}{2\sin\text{x}+\cos\text{x}+3}\ \text{dx}$
✓
Answer
Let $\text{I}=\int\frac{3+2\cos\text{x}+4\sin\text{x}}{2\sin\text{x}+\cos\text{x}+3}\ \text{dx}$
Let $3+2\cos\text{x}+4\sin\text{x}=\lambda\frac{\text{d}}{\text{dx}}(2\sin\text{x}+\cos\text{x}+3)+\mu(2\sin\text{x}+\cos\text{x}+3)+\text{v}$
$3+2\cos\text{x}+4\sin\text{x}=\lambda(2\cos\text{x}-\sin\text{x})+\mu(2\sin\text{x}+\cos\text{x}+3)+\text{v}$
$3+2\cos\text{x}+4\sin\text{x}=(-\lambda+2\mu)\sin\text{x}+(2\lambda+\mu)\cos\text{x}+3\mu+\text{v}$
Comparing the coefficient of $\sin\text{x}\ \&\cos\text{x}$ on the both the sides,
$-\lambda+2\mu\ =4\dots\dots(1)$
$2\lambda + \mu = 2 \ \dots\dots(2)$
$2\mu + \text{v} = 3 \ \dots\dots(3)$
Solving Equation (1), (2) and (3), we get
$\lambda=0,\mu=2,\text{v}=-3$
$\text{I}==\int\frac{2(2\sin\text{x}+\cos\text{x}+3)-3}{(2\sin\text{x}+\cos\text{x}+3)}\ \text{dx}$
$=2\int\text{dx}-2\int\frac{1}{2\sin\text{x}+\cos\text{x}+3}\ \text{dx}$
$\text{I}=2\text{x}-3\text{I}_1+\text{C}_1\dots\dots(4)$
Let $\text{I}_1=\int\frac{1}{2\sin\text{x}+\cos\text{x}+3}\ \text{dx}$
$\sin\text{x}=\frac{2\tan\frac{\text{x}}{2}}{\Big(1+\tan^2\frac{\text{x}}{2}\Big)},\cos\text{x}=\frac{1-\tan^2\frac{\text{x}}{2}}{\Big(1+\tan^2\frac{\text{x}}{2}\Big)}$
$\text{I}_1=\int\frac{1}{2\Bigg(\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)+\Bigg(\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)+3}\ \text{dx}$
$=\int\frac{\Big(1+\tan^2\frac{\text{x}}{2}\Big)}{4\tan\frac{\text{x}}{2}+1-\tan^2\frac{\text{x}}{2}+3\Big(1+\tan^2\frac{\text{x}}{2}\Big)}\ \text{dx}$
$=\int\frac{\sec^2\frac{\text{x}}{2}}{2\tan^2\frac{\text{x}}{2}+4\tan\frac{\text{x}}{2}+4}\ \text{dx}$
Let $\tan\frac{\text{x}}{2}=\text{t}$
$\frac{1}{2}\sec^2\frac{\text{x}}{2}=\text{dt}$
$\text{I}_1=\int\frac{2\text{dt}}{2\text{t}^2+4\text{t}+4}$
$\frac{2}{2}\int\frac{\text{dt}}{\text{t}^2+2\text{t}+2}$
$=\int\frac{\text{dt}}{\text{t}^2+2\text{t}+1-1+2}$
$=\int\frac{\text{dt}}{(\text{t}+1)^2+1}$
$=\tan^{-1}\Big(\tan\frac{\text{x}}{2}+1)+\text{C}_2$
Now, using equation (1),
$\text{I}=2\times-3\tan^{-1}\Big(\tan\frac{\text{x}}{2}+1\Big)+\text{C}$
Let $3+2\cos\text{x}+4\sin\text{x}=\lambda\frac{\text{d}}{\text{dx}}(2\sin\text{x}+\cos\text{x}+3)+\mu(2\sin\text{x}+\cos\text{x}+3)+\text{v}$
$3+2\cos\text{x}+4\sin\text{x}=\lambda(2\cos\text{x}-\sin\text{x})+\mu(2\sin\text{x}+\cos\text{x}+3)+\text{v}$
$3+2\cos\text{x}+4\sin\text{x}=(-\lambda+2\mu)\sin\text{x}+(2\lambda+\mu)\cos\text{x}+3\mu+\text{v}$
Comparing the coefficient of $\sin\text{x}\ \&\cos\text{x}$ on the both the sides,
$-\lambda+2\mu\ =4\dots\dots(1)$
$2\lambda + \mu = 2 \ \dots\dots(2)$
$2\mu + \text{v} = 3 \ \dots\dots(3)$
Solving Equation (1), (2) and (3), we get
$\lambda=0,\mu=2,\text{v}=-3$
$\text{I}==\int\frac{2(2\sin\text{x}+\cos\text{x}+3)-3}{(2\sin\text{x}+\cos\text{x}+3)}\ \text{dx}$
$=2\int\text{dx}-2\int\frac{1}{2\sin\text{x}+\cos\text{x}+3}\ \text{dx}$
$\text{I}=2\text{x}-3\text{I}_1+\text{C}_1\dots\dots(4)$
Let $\text{I}_1=\int\frac{1}{2\sin\text{x}+\cos\text{x}+3}\ \text{dx}$
$\sin\text{x}=\frac{2\tan\frac{\text{x}}{2}}{\Big(1+\tan^2\frac{\text{x}}{2}\Big)},\cos\text{x}=\frac{1-\tan^2\frac{\text{x}}{2}}{\Big(1+\tan^2\frac{\text{x}}{2}\Big)}$
$\text{I}_1=\int\frac{1}{2\Bigg(\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)+\Bigg(\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)+3}\ \text{dx}$
$=\int\frac{\Big(1+\tan^2\frac{\text{x}}{2}\Big)}{4\tan\frac{\text{x}}{2}+1-\tan^2\frac{\text{x}}{2}+3\Big(1+\tan^2\frac{\text{x}}{2}\Big)}\ \text{dx}$
$=\int\frac{\sec^2\frac{\text{x}}{2}}{2\tan^2\frac{\text{x}}{2}+4\tan\frac{\text{x}}{2}+4}\ \text{dx}$
Let $\tan\frac{\text{x}}{2}=\text{t}$
$\frac{1}{2}\sec^2\frac{\text{x}}{2}=\text{dt}$
$\text{I}_1=\int\frac{2\text{dt}}{2\text{t}^2+4\text{t}+4}$
$\frac{2}{2}\int\frac{\text{dt}}{\text{t}^2+2\text{t}+2}$
$=\int\frac{\text{dt}}{\text{t}^2+2\text{t}+1-1+2}$
$=\int\frac{\text{dt}}{(\text{t}+1)^2+1}$
$=\tan^{-1}\Big(\tan\frac{\text{x}}{2}+1)+\text{C}_2$
Now, using equation (1),
$\text{I}=2\times-3\tan^{-1}\Big(\tan\frac{\text{x}}{2}+1\Big)+\text{C}$
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