Answer

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**Hint:**To solve this integral, we will rewrite the numerator in such a manner that we can simplify the integrand. We will also use the method of substitution for the purpose of simplification. We will need the following integration formulae of standard equations, $\int{\dfrac{1}{x}dx=\ln x+c}$ and $\int{1dx}=x+c$ where $c$ is the integration constant. Using these, we will be able to evaluate the given integral.

**Complete step-by-step solution:**The given integral is $I=\int{\dfrac{2\sin \theta +\cos \theta }{7\sin \theta -5\cos \theta }d\theta }$. The denominator is $7\sin \theta -5\cos \theta $ and the numerator is $2\sin \theta +\cos \theta $. It will become easier to find the value of the integration if we simplify the integrand. Looking at the integrand, we can aim to write the numerator in terms of the denominator. Now, the coefficients of the sine and cosine functions in the numerator is 2 and 1 respectively. The numerator of the integrand can be modified and rewritten in the following manner,

$2\sin \theta +\cos \theta =\dfrac{148}{74}\sin \theta +\dfrac{74}{74}\cos \theta $

The number 74 is works in this case because $74=49+25$ which is nothing but $74={{7}^{2}}+{{5}^{2}}$ which are the coefficients of the sine and cosine function in the denominator.

We will split the right hand side of the above equation so that we have the denominator as a part of the numerator expression, as follows,

$\begin{align}

& \dfrac{148}{74}\sin \theta +\dfrac{74}{74}\cos \theta =\dfrac{63+85}{74}\sin \theta +\dfrac{119-45}{74}\cos \theta \\

& \Rightarrow \dfrac{148}{74}\sin \theta +\dfrac{74}{74}\cos \theta =\left( \dfrac{63}{74}\sin \theta -\dfrac{45}{74}\cos \theta \right)+\left( \dfrac{85}{74}\sin \theta +\dfrac{119}{74}\cos \theta \right) \\

& \therefore \dfrac{148}{74}\sin \theta +\dfrac{74}{74}\cos \theta =\dfrac{9}{74}\left( 7\sin \theta -5\cos \theta \right)+\dfrac{17}{74}\left( 5\sin \theta +7\cos \theta \right) \\

\end{align}$

Now, substituting this expression in place of the numerator in the integrand, we get

$\begin{align}

& I=\int{\dfrac{\dfrac{9}{74}\left( 7\sin \theta -5\cos \theta \right)+\dfrac{17}{74}\left( 5\sin \theta +7\cos \theta \right)}{7\sin \theta -5\cos \theta }}d\theta \\

& =\dfrac{9}{74}\int{\dfrac{\left( 7\sin \theta -5\cos \theta \right)}{7\sin \theta -5\cos \theta }d\theta +\dfrac{17}{74}\int{\dfrac{\left( 5\sin \theta +7\cos \theta \right)}{7\sin \theta -5\cos \theta }}}d\theta

\end{align}$

Now, let us integrate the first term of $I$.

$\dfrac{9}{74}\int{\dfrac{\left( 7\sin \theta -5\cos \theta \right)}{7\sin \theta -5\cos \theta }d\theta =\dfrac{9}{74}\int{1d\theta }}$

We know that $\int{1dx}=x+c$. Therefore, the first term of $I$ equals to $\dfrac{9}{74}\theta +{{c}_{1}}$.

Now, for the second term of $I$, we will use the method of substitution. Let $u=7\sin \theta -5\cos \theta $. Therefore, on differentiating $u$ we have the following,

$\begin{align}

& \dfrac{du}{d\theta }=7\cos \theta +5\sin \theta \\

& \therefore du=\left( 7\cos \theta +5\sin \theta \right)d\theta \\

\end{align}$

Therefore, the second term of $I$ will become $\dfrac{17}{74}\int{\dfrac{1}{u}du}$. Now, we know that $\int{\dfrac{1}{x}dx=\ln x+c}$.

So, the second term of $I$ equals to $\dfrac{17}{74}\ln \left( u \right)=\dfrac{17}{74}\ln \left( 7\sin \theta -5\cos \theta \right)+{{c}_{2}}$.

Therefore, we have the following equation,

$\begin{align}

& I=\dfrac{9}{74}\theta +{{c}_{1}}+\dfrac{17}{74}\ln \left( 7\sin \theta -5\cos \theta \right)+{{c}_{2}} \\

& =\dfrac{9}{74}\theta +\dfrac{17}{74}\ln \left( 7\sin \theta -5\cos \theta \right)+C

\end{align}$

**Note:**We were able to rewrite the numerator of the integrand in such a way that the expression in the denominator was a part of the modified numerator. Also, the sine and cosine functions are related to each other in integration and differentiation. Therefore, we can use the substitution method with convenience for evaluating this integral. The calculations and substitutions need to be done carefully so that we can avoid making any minor mistakes.

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