The trigonometric functions $\sin{x}$ and $\cos{x}$ formed a rational expression and the definite integral of this function has to evaluate from $0$ to $\dfrac{\pi}{2}$ in this definite integral problem.
$\displaystyle \int \limits_0^{\Large \frac{\pi}{2}} {\dfrac{\cos{x}}{6-5\sin{x}+\sin^2{x}} \,}dx$
In the given trigonometric rational expression, there is a quadratic expression in terms of sine function in denominator and the derivative of the sine function is appearing in the numerator of the same function. So, the rational expression can be transformed into algebraic expression by differentiating the sine function.
Let $u = \sin{x}$. Differentiate the equation with respect to $x$.
$\implies$ $\dfrac{du}{dx} \,=\, \dfrac{d}{dx}{\, \sin{x}}$
$\implies$ $\dfrac{du}{dx} \,=\, \cos{x}$
$\implies$ $du \,=\, \cos{x} \times dx$
Now, simplify the rational expression to convert the function in terms of $u$.
$=\,\,\,$ $\displaystyle \int \limits_0^{\Large \frac{\pi}{2}} {\dfrac{\cos{x}}{6-5\sin{x}+\sin^2{x}} \,}dx$
$=\,\,\,$ $\displaystyle \int \limits_0^{\Large \frac{\pi}{2}} {\dfrac{1 \times \cos{x}}{6-5\sin{x}+\sin^2{x}} \,}dx$
$=\,\,\,$ $\displaystyle \int \limits_0^{\Large \frac{\pi}{2}} {\dfrac{1}{6-5\sin{x}+\sin^2{x}} \,}(\cos{x} \times dx)$
The lower and upper limits are $0$ and $\dfrac{\pi}{2}$ but they are the limits of the rational function when the function is in terms of $x$. Actually, we are about to convert it into $u$. So, we must also convert the lower and upper limits.
$(1)\,\,\,$ When $x = 0$, then $u = \sin{(0)} = 0$.
$(2)\,\,\,$ When $x = \dfrac{\pi}{2}$, then $u = \sin{\Big(\dfrac{\pi}{2}\Big)} = 1$.
$\implies$ $\displaystyle \int \limits_0^{\Large \frac{\pi}{2}} {\dfrac{1}{6-5\sin{x}+\sin^2{x}} \,}(\cos{x} \times dx)$ $\,=\,$ $\displaystyle \int \limits_0^1 {\dfrac{1}{6-5u+u^2} \,}du$
Now, continue the simplifying the rational expression in algebraic form.
$=\,\,\,$ $\displaystyle \int \limits_0^1 {\dfrac{1}{u^2-5u+6} \,}du$
$=\,\,\,$ $\displaystyle \int \limits_0^1 {\dfrac{1}{u^2-3u-2u+6} \,}du$
$=\,\,\,$ $\displaystyle \int \limits_0^1 {\dfrac{1}{u(u-3)-2(u-3)} \,}du$
$=\,\,\,$ $\displaystyle \int \limits_0^1 {\dfrac{1}{(u-3)(u-2)} \,}du$
The algebraic expression represents a rational expression that consists of non-repeated linear factors in denominator. It can be decomposed as the sum of the partial fractions.
$\dfrac{1}{(u-3)(u-2)}$ $\,=\,$ $\dfrac{A}{u-3}$ $+$ $\dfrac{B}{u-2}$
Substitute $u = 3$, then $A$ $\,=\,$ $\dfrac{1}{3-2}$ $\,=\,$ $\dfrac{1}{1}$ $\,=\,$ $1$
Substitute $u = 2$, then $B$ $\,=\,$ $\dfrac{1}{2-3}$ $\,=\,$ $\dfrac{1}{-1}$ $\,=\,$ $-1$
$\implies$ $\dfrac{1}{(u-3)(u-2)}$ $\,=\,$ $\dfrac{1}{u-3}$ $+$ $\dfrac{-1}{u-2}$
$\implies$ $\dfrac{1}{(u-3)(u-2)}$ $\,=\,$ $\dfrac{1}{u-3}$ $-$ $\dfrac{1}{u-2}$
$\implies$ $\displaystyle \int \limits_0^1 {\dfrac{1}{(u-3)(u-2)} \,}du$ $\,=\,$ $\displaystyle \int \limits_0^1 {\Bigg(\dfrac{1}{u-3}-\dfrac{1}{u-2}\Bigg) \,}du$
$\implies$ $\displaystyle \int \limits_0^1 {\Bigg(\dfrac{1}{u-3}-\dfrac{1}{u-2}\Bigg) \,}du$ $\,=\,$ $\displaystyle \int \limits_0^1 {\dfrac{1}{u-3}\,}du$ $-$ $\displaystyle \int \limits_0^1 {\dfrac{1}{u-2} \,}du$
The denominator is a linear expression in one variable in each integral term.
$=\,\,\,$ $\displaystyle \int \limits_0^1 {\dfrac{1}{u-3}\,}du$ $-$ $\displaystyle \int \limits_0^1 {\dfrac{1}{u-2} \,}du$
So, use integral rule for reciprocal of linear expression in one variable to evaluate integration of every rational expression.
$=\,\,\,$ $\dfrac{1}{1}\Big[\log_e{|u-3|}+c_1\Big]_0^1$ $-$ $\dfrac{1}{1}\Big[\log_e{|u-2|}+c_2\Big]_0^1$
$=\,\,\,$ $1 \times \Big[\log_e{|u-3|}+c_1\Big]_0^1$ $-$ $1 \times \Big[\log_e{|u-2|}+c_2\Big]_0^1$
$=\,\,\,$ $\Big[\log_e{|u-3|}+c_1\Big]_0^1$ $-$ $\Big[\log_e{|u-2|}+c_2\Big]_0^1$
$=\,\,\,$ $\Big[(\log_e{|1-3|}+c_1)$ $-$ $(\log_e{|0-3|}+c_1)\Big]$ $-$ $\Big[(\log_e{|1-2|}+c_2)$ $-$ $(\log_e{|0-2|}+c_2)\Big]$
$=\,\,\,$ $\Big[\log_e{|-2|}+c_1$ $-$ $\log_e{|-3|}-c_1\Big]$ $-$ $\Big[\log_e{|-1|}+c_2$ $-$ $\log_e{|-2|}-c_2\Big]$
$=\,\,\,$ $\require{cancel} \Big[\log_e{|-2|}+\cancel{c_1}$ $-$ $\require{cancel} \log_e{|-3|}-\cancel{c_1}\Big]$ $-$ $\require{cancel} \Big[\log_e{|-1|}+\cancel{c_2}$ $-$ $\require{cancel} \log_e{|-2|}-\cancel{c_2}\Big]$
$=\,\,\,$ $\Big[\log_e{|-2|}-\log_e{|-3|}\Big]$ $-$ $\Big[\log_e{|-1|}-\log_e{|-2|}\Big]$
$=\,\,\,$ $\log_e{\Bigg|\dfrac{-2}{-3}\Bigg|}$ $-$ $\log_e{\Bigg|\dfrac{-1}{-2}\Bigg|}$
$=\,\,\,$ $\log_e{\Bigg|\dfrac{2}{3}\Bigg|}$ $-$ $\log_e{\Bigg|\dfrac{1}{2}\Bigg|}$
$=\,\,\,$ $\log_e{\Bigg|\dfrac{2}{3}\Bigg|}$ $-$ $\log_e{\Bigg|\dfrac{1}{2}\Bigg|}$
$=\,\,\,$ $\log_e{\Bigg|\dfrac{2}{3}\Big|}$ $-$ $\log_e{\Bigg|2^{-1}\Big|}$
$=\,\,\,$ $\log_e{\Bigg|\dfrac{2}{3}\Bigg|}$ $+$ $\log_e{|2|}$
$=\,\,\,$ $\log_e{\Bigg|\dfrac{2}{3} \times 2\Bigg|}$
$=\,\,\,$ $\log_e{\Bigg|\dfrac{2 \times 2}{3}\Bigg|}$
$=\,\,\,$ $\log_e{\Bigg|\dfrac{4}{3}\Bigg|}$
Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.