Question

Evaluate the integral of $\dfrac{\cos x}{\left( 1-\sin x \right)\left( 2-\sin x \right)}$.

Answer 1

Hint: The integral that we are given in the problem is $\dfrac{\cos x}{\left( 1-\sin x \right)\left( 2-\sin x \right)}$ . First, let us assume the expression $1-\sin x=t$ . Taking differentials on both sides of the equation, we get $-\cos x dx=dt$ . The integral thus becomes $\Rightarrow I=\int{\dfrac{-dt}{t\left( t+1 \right)}}$ . Using partial fraction technique, we get $\Rightarrow I=-\int{\left( \dfrac{1}{t}-\dfrac{1}{t+1} \right)}dt$ . Carrying out the integration, we get $\Rightarrow I=-\ln t+\ln \left( t+1 \right)$ . Upon substituting the value of t, we get the required result.

Complete step-by-step solution:
Let the integral be,
$I=\int{\dfrac{\cos x}{\left( 1-\sin x \right)\left( 2-\sin x \right)}dx}$
Now, let us assume the expression $1-\sin x=t$ . Taking differentials on both sides of the equation, we get,
$-\cos xdx=dt$
So, incorporating this in the integral, the integral thus becomes,
 $\Rightarrow I=\int{\dfrac{-dt}{t\left( t+1 \right)}}$
Now, we take the help of partial fractions. For this, we need to break the expression $\dfrac{1}{t\left( t+1 \right)}$ into two or more simpler fractions. We do so by assuming,
$\dfrac{1}{t\left( t+1 \right)}=\dfrac{A}{t}+\dfrac{B}{t+1}$
Multiplying $t\left( t+1 \right)$ on both sides of the above equation, we get,
$\Rightarrow 1=A\left( t+1 \right)+Bt$
Simplifying the above equation, we get,
$\Rightarrow 1=A+\left( A+B \right)t$
Comparing the LHS and the RHS, we get,
$\begin{align}
  & A=1,A+B=0 \\
 & \Rightarrow A=1,B=-1 \\
\end{align}$
So, we can write the expression $\dfrac{1}{t\left( t+1 \right)}$ as $\dfrac{1}{t\left( t+1 \right)}=\dfrac{1}{t}-\dfrac{1}{t+1}$ . The above integral thus becomes,
$\Rightarrow I=-\int{\left( \dfrac{1}{t}-\dfrac{1}{t+1} \right)}dt$
Opening up the brackets, we get,
$\Rightarrow I=-\int{\dfrac{1}{t}}dt+\int{\dfrac{1}{t+1}dt}$
The above integral can be rewritten as,
$\Rightarrow I=-\int{\dfrac{1}{t}}dt+\int{\dfrac{d\left( t+1 \right)}{t+1}}$
Carrying out the integration, we get,
$\Rightarrow I=-\ln t+\ln \left( t+1 \right)$
Including the constant of integration as the integration is indefinite, we get,
$\Rightarrow I=-\ln t+\ln \left( t+1 \right)+c$
Upon substituting the value of t as $t=1-\sin x$ , we get,
$\Rightarrow I=-\ln \left( 1-\operatorname{sinx} \right)+\ln \left( 2-\sin x \right)+c$
Thus, we can conclude that the given integral $\dfrac{\cos x}{\left( 1-\sin x \right)\left( 2-\sin x \right)}$ upon integration gives $-\ln \left( 1-\operatorname{sinx} \right)+\ln \left( 2-\sin x \right)+c$ which is option B.

Note: The problem can be solved in another way. The integral is $I=\int{\dfrac{\cos x}{\left( 1-\sin x \right)\left( 2-\sin x \right)}dx}$ . We can rewrite the integral as,
$I=\int{\cos x\left( \dfrac{1}{\left( 1-\sin x \right)\left( 2-\sin x \right)} \right)dx}$
Now, using partial fractions, we get,
\[\begin{align}
  & I=\int{\cos x\left( \dfrac{1}{\left( 1-\sin x \right)}-\dfrac{1}{\left( 2-\sin x \right)} \right)dx} \\
 & \Rightarrow I=\int{\dfrac{\cos xdx}{\left( 1-\sin x \right)}}-\int{\dfrac{\cos xdx}{\left( 2-\sin x \right)}} \\
\end{align}\]
The integral can be modified as \[\Rightarrow I=-\int{\dfrac{d\left( 1-\sin x \right)}{\left( 1-\sin x \right)}}+\int{\dfrac{d\left( 2-\sin x \right)}{\left( 2-\sin x \right)}}\] . This gives, $\Rightarrow I=-\ln \left( 1-\operatorname{sinx} \right)+\ln \left( 2-\sin x \right)+c$ .