Evaluate the following intregals: (1- )(2- ) dx - Vidyadip

Question

Evaluate the following intregals:
$\int\frac{\cos\text{x}}{(1-\sin\text{x})(2-\sin\text{x})}\ \text{dx}$

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Answer

We have,
$\text{I}=\int\frac{\cos\text{x}\text{ dx}}{(1-\sin\text{x})(2-\sin\text{x})}$
putting $\sin\text{x}=\text{t}$
$\Rightarrow\cos\text{x dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{(1-\text{t})(2-\text{t})}$
$\therefore\text{I}=\int\frac{\text{dt}}{(1-\text{t})(2-\text{t})}$
Let $\frac{1}{(\text{t}-1)(\text{t}-2)}=\frac{\text{A}}{\text{t}-1}+\frac{ \text{B}}{\text{t}-2}$
$\Rightarrow\frac{1}{(\text{t}-1)(\text{t}-2)}=\frac{\text{A}(\text{t}-2)+\text{B}(\text{t}-1)}{(\text{t}-1)(\text{t}-2)}$
$\Rightarrow1=\text{A}(\text{t}-2)+\text{B}(\text{t}-1)$
Putting t - 1 = 0
⇒ t = 1
$\therefore$ 1 = A (1 - 2) + B × 0
⇒ A = -1
Putting t - 2 = 0
⇒ t = 2
$\therefore$ 1 = A × 0 + B(2 - 1)
⇒ B = 1
$\therefore\text{I}=\int\frac{-\text{dt}}{\text{t}-1}+\int\frac{\text{dt}}{\text{t}-2}$
$=-\log|\text{t}-1|+\log|\text{t}-2|+\text{C}$
$=\log\Big|\frac{\text{t}-2}{\text{t}-1}\Big|+\text{C}$
$=\log\Big|\frac{\sin\text{x}-2}{\sin\text{x}-1}\Big|+\text{C}$
$=\log\Big|\frac{2-\sin\text{x}}{1-\sin\text{x}}\Big|+\text{C}$

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