Evaluate the following integrals: 1 + 2x dx - Vidyadip

Question

Evaluate the following integrals:
$\int\frac{1}{\sin\text{x}+\sin2\text{x}}\ \text{dx}$

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Answer

Let $\text{I}=\int\frac{1}{\sin\text{x}+\sin2\text{x}}\ \text{dx}$
$=\int\frac{\text{dx}}{\sin\text{x}+2\sin\text{x}\cos\text{x}}$
$=\int\frac{\sin\text{x dx}}{(1-\cos^2\text{x})+2(1-\cos^2\text{x})\cos\text{x}}$
Let $\cos\text{x}=\text{t}\Rightarrow-\sin\text{x dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{(\text{t}^2-1)+2(\text{t}^2-1)\text{t}} $
$=\int\frac{\text{dt}}{(\text{t}^2-1)(1+2\text{t})}$
Let $\int\frac{1}{(\text{t}^2-1)(1+2\text{t})}=\frac{\text{A}}{\text{t}-1}+\frac{\text{B}}{\text{t}+1}+\frac{\text{C}}{1+2\text{t}}$
Put t = 1
⇒ 1 = 6A $\Rightarrow\text{A}=\frac{1}{6}$
Put t = -1
⇒ 1 = 2B $\Rightarrow\text{B}=\frac{1}{2}$
put $\text{t}=-\frac{1}{2}$
$\Rightarrow1=-\frac{3}{4}\text{C}\Rightarrow\text{C}=-\frac{4}{3}$
Thus,
$\text{I}=\frac{1}{6}\int\frac{\text{dt}}{\text{t}-1}+\frac{1}{2}\int\frac{\text{dt}}{\text{t}+1}-\frac{4}{3}\int\frac{\text{dt}}{1+2\text{t}}$
$=\frac{1}{6}\log|\text{t}-1|+\frac{1}{2}\log|\text{t}+1|-\frac{2}{3}\log|1+2\text{t}|+\text{C}$
Hence,
$\text{I}=\frac{1}{6}\log|\cos\text{x}-1|+\frac{1}{2}\log|\cos\text{x}+1|\\-\frac{2}{3}\log|1+2\cos\text{x}|+\text{C}$

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