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

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

Explanation

Solution

The integral that we are given in the problem is cos⁡x(1−sin⁡x)(2−sin⁡x)\dfrac{\cos x}{\left( 1-\sin x \right)\left( 2-\sin x \right)} . First, let us assume the expression 1−sin⁡x=t1-\sin x=t . Taking differentials on both sides of the equation, we get −cos⁡xdx=dt-\cos x dx=dt . The integral thus becomes ⇒I=∫−dtt(t+1)\Rightarrow I=\int{\dfrac{-dt}{t\left( t+1 \right)}} . Using partial fraction technique, we get ⇒I=−∫(1t−1t+1)dt\Rightarrow I=-\int{\left( \dfrac{1}{t}-\dfrac{1}{t+1} \right)}dt . Carrying out the integration, we get ⇒I=−ln⁡t+ln⁡(t+1)\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=∫cos⁡x(1−sin⁡x)(2−sin⁡x)dxI=\int{\dfrac{\cos x}{\left( 1-\sin x \right)\left( 2-\sin x \right)}dx}
Now, let us assume the expression 1−sin⁡x=t1-\sin x=t . Taking differentials on both sides of the equation, we get,
−cos⁡xdx=dt-\cos xdx=dt
So, incorporating this in the integral, the integral thus becomes,
⇒I=∫−dtt(t+1)\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 1t(t+1)\dfrac{1}{t\left( t+1 \right)} into two or more simpler fractions. We do so by assuming,
1t(t+1)=At+Bt+1\dfrac{1}{t\left( t+1 \right)}=\dfrac{A}{t}+\dfrac{B}{t+1}
Multiplying t(t+1)t\left( t+1 \right) on both sides of the above equation, we get,
⇒1=A(t+1)+Bt\Rightarrow 1=A\left( t+1 \right)+Bt
Simplifying the above equation, we get,
⇒1=A+(A+B)t\Rightarrow 1=A+\left( A+B \right)t
Comparing the LHS and the RHS, we get,
A=1,A+B=0 ⇒A=1,B=−1 \begin{aligned} & A=1,A+B=0 \\\ & \Rightarrow A=1,B=-1 \\\ \end{aligned}
So, we can write the expression 1t(t+1)\dfrac{1}{t\left( t+1 \right)} as 1t(t+1)=1t−1t+1\dfrac{1}{t\left( t+1 \right)}=\dfrac{1}{t}-\dfrac{1}{t+1} . The above integral thus becomes,
⇒I=−∫(1t−1t+1)dt\Rightarrow I=-\int{\left( \dfrac{1}{t}-\dfrac{1}{t+1} \right)}dt
Opening up the brackets, we get,
⇒I=−∫1tdt+∫1t+1dt\Rightarrow I=-\int{\dfrac{1}{t}}dt+\int{\dfrac{1}{t+1}dt}
The above integral can be rewritten as,
⇒I=−∫1tdt+∫d(t+1)t+1\Rightarrow I=-\int{\dfrac{1}{t}}dt+\int{\dfrac{d\left( t+1 \right)}{t+1}}
Carrying out the integration, we get,
⇒I=−ln⁡t+ln⁡(t+1)\Rightarrow I=-\ln t+\ln \left( t+1 \right)
Including the constant of integration as the integration is indefinite, we get,
⇒I=−ln⁡t+ln⁡(t+1)+c\Rightarrow I=-\ln t+\ln \left( t+1 \right)+c
Upon substituting the value of t as t=1−sin⁡xt=1-\sin x , we get,
⇒I=−ln⁡(1−sinx⁡)+ln⁡(2−sin⁡x)+c\Rightarrow I=-\ln \left( 1-\operatorname{sinx} \right)+\ln \left( 2-\sin x \right)+c
Thus, we can conclude that the given integral cos⁡x(1−sin⁡x)(2−sin⁡x)\dfrac{\cos x}{\left( 1-\sin x \right)\left( 2-\sin x \right)} upon integration gives −ln⁡(1−sinx⁡)+ln⁡(2−sin⁡x)+c-\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=∫cos⁡x(1−sin⁡x)(2−sin⁡x)dxI=\int{\dfrac{\cos x}{\left( 1-\sin x \right)\left( 2-\sin x \right)}dx} . We can rewrite the integral as,
I=∫cos⁡x(1(1−sin⁡x)(2−sin⁡x))dxI=\int{\cos x\left( \dfrac{1}{\left( 1-\sin x \right)\left( 2-\sin x \right)} \right)dx}
Now, using partial fractions, we get,

& 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{aligned}$$ The integral can be modified as $$\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, $\Rightarrow I=-\ln \left( 1-\operatorname{sinx} \right)+\ln \left( 2-\sin x \right)+c$ .