Evaluate the following integrals: 1 (5-4 ) dx - Vidyadip

Question

Evaluate the following integrals:
$\int\frac{1}{\cos\text{x}(5-4\sin\text{x})}\ \text{dx}$

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Answer

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