Question

Integrate $\dfrac{{\cos x}}{{(1 - \sin x)(2 - \sin x)}}$.

Answer 1

Here we have to integrate the given function. The function is in the form of a fraction of the trigonometric function. Firstly we substitute the $\sin x$ as t. Since it is in the form of fraction, we use the integration by partial fractions and by integration formula we are going to obtain the answer for the given question.

Complete step by step answer:
Now consider the given question
$\Rightarrow \int {\dfrac{{\cos x}}{{(1 - \sin x)(2 - \sin x)}}} .dx$
Substitute $\sin x = t$, on differentiating we have $\cos x\,dx = dt$
$\Rightarrow \int {\dfrac{{dt}}{{(1 - t)(2 - t)}}}$-------(1)
We express the fractions by partial fractions. So we have
$\Rightarrow \dfrac{1}{{(1 - t)(2 - t)}} = \dfrac{A}{{(1 - t)}} + \dfrac{B}{{(2 - t)}}$-------(2)
On taking LCM for the RHS we have
$\Rightarrow \dfrac{1}{{(1 - t)(2 - t)}} = \dfrac{{A(2 - t) + B(1 - t)}}{{(1 - t)(2 - t)}}$
Cancelling the terms which are present in the denominator.
$\Rightarrow 1 = A(2 - t) + B(1 - t)$ ----(3)

By considering the above equation we have to determine the value of A and B.
Take $t = 1$ in the equation (3), we have
$\Rightarrow 1 = A(2 - 1) + B(1 - 1)$
On simplifying we have
$\Rightarrow 1 = A$------(4)
Therefore the value of A is 1
Now we determine the value of B by considering $t = 2$ in equation (3)
$\Rightarrow 1 = A(2 - 2) + B(1 - 2)$
On simplifying we have
$\Rightarrow 1 = B( - 1)$
$\Rightarrow B = - 1$ ----- (5)

On substituting the equation (4) and the equation (5) in the equation (2)
$\Rightarrow \dfrac{1}{{(1 - t)(2 - t)}} = \dfrac{1}{{(1 - t)}} + \dfrac{{ - 1}}{{(2 - t)}}$
By using the sign conventions, we have
$\Rightarrow \dfrac{1}{{(1 - t)(2 - t)}} = \dfrac{1}{{(1 - t)}} - \dfrac{1}{{(2 - t)}}$
Therefore, the equation (1) can be written as
$\Rightarrow \int {\dfrac{{dt}}{{(1 - t)(2 - t)}}} = \int {\left( {\dfrac{1}{{(1 - t)}} - \dfrac{1}{{(2 - t)}}} \right)} \,dt$
By the property of integration $\int {(a - b)\,dx = \int {a.dx - \int {b.dx} } }$, so the above inequality is written as
$\Rightarrow \int {\dfrac{{dt}}{{(1 - t)(2 - t)}}} = \int {\dfrac{{dt}}{{(1 - t)}}} - \int {\dfrac{{dt}}{{(2 - t)}}}$

As we know the integration formula $\int {\dfrac{1}{x}dx} = \log |x|$. The above inequality resembles the same integration formula. On applying the formula, we have
$\Rightarrow \int {\dfrac{{dt}}{{(1 - t)(2 - t)}}} = \log |1 - t| - \log |2 - t| + C$
On using the property of logarithmic function $\log m - \log n = \log \left( {\dfrac{m}{n}} \right)$, so the above inequality is written as
$\Rightarrow \int {\dfrac{{dt}}{{(1 - t)(2 - t)}}} = \log \left( {\dfrac{{|1 - t|}}{{|2 - t|}}} \right) + C$
On substituting the value of t in the above inequality we have
$\Rightarrow \int {\dfrac{{\cos x}}{{(1 - \sin x)(2 - \sin x)}}dx} = \log \left( {\dfrac{{|1 - \sin x|}}{{|2 - \sin x|}}} \right) + C$
where $C$ is integration constant.

Therefore, the integration of $\dfrac{{\cos x}}{{(1 - \sin x)(2 - \sin x)}}$ is $\log \left( {\dfrac{{|1 - \sin x|}}{{|2 - \sin x|}}} \right) + C$.

Note: An algebraic fraction can be broken down into simpler parts known as partial fractions. The important partial fractions form of the rational fraction is given by
1. $\dfrac{{px + q}}{{(x - a)(x - b)}},a \ne b\, \to \,\dfrac{A}{{(x - a)}} + \dfrac{B}{{(x - b)}}$
2. $\dfrac{{px + q}}{{{{(x - a)}^2}}}\, \to \,\dfrac{A}{{(x - a)}} + \dfrac{B}{{{{(x - a)}^2}}}$
3. $\dfrac{{p{x^2} + qx + r}}{{(x - a)(x - b)(x - c)}}\, \to \,\dfrac{A}{{(x - a)}} + \dfrac{B}{{(x - b)}} + \dfrac{C}{{(x - c)}}$
4. $\dfrac{{p{x^2} + qx + r}}{{{{(x - a)}^2}(x - b)}}\, \to \,\dfrac{A}{{(x - a)}} + \dfrac{B}{{{{(x - a)}^2}}} + \dfrac{C}{{(x - b)}}$
5. $\dfrac{{p{x^2} + qx + r}}{{(x - a)({x^2} + bx + c)}}\, \to \,\dfrac{A}{{(x - a)}} + \dfrac{{Bx + C}}{{({x^2} + bx + c)}}$
The way of considering the partial fractions will differ and it is based on the form of the rational fraction.