91Ó°ÊÓ

To evaluate the first integral, we should follow these steps: Step 1: Apply substitution Make a substitution for \(\cos x+\sin x\) where \(u = \cos x+\sin x\). Differentiate \(u\) with respect to \(x\) to get the differential form: \begin{align} \dfrac{d u}{d x}= -\sin x+\cos x d x. \end{align} Step 2: Rearrange and write terms in terms of \(u\) Re-write the integral in terms of \(u\): \begin{align} \int \dfrac{\cos x}{u} (-\sin x +\cos x ) d x = -\int \dfrac{1}{u} du. \end{align} Step 3: Evaluate the integral Now, evaluate the integral using the rule \(\int \dfrac{1}{u} du =\ln{|u|}+C\) where C is the constant of integration: \begin{align} -\int \dfrac{1}{u} du = -\ln{|u|}+C = -\ln{|\cos x+\sin x|}+C. \end{align} Step 4: Write the final answer Our final answer is: \begin{align} \int \dfrac{\cos x}{\cos x+\sin x} d x = -\ln{|\cos x+\sin x|}+C. \end{align}

To evaluate the second integral, we should follow these steps: Step 1: Apply substitution Make a substitution for \(\cot x\) where \(u = \cot x\). So, \(\frac{du}{dx} = -\csc^2 x dx\). And, \(\csc^2x = \frac{1}{\sin^2 x}\). So, \(\frac{du}{dx} = -\frac{1}{\sin^2x} dx\). Step 2: Rewrite the integral in terms of \(u\) Re-write the integral in terms of \(u\): \begin{align} \int \frac{1}{a+b u} \left(-\frac{1}{\sin^2x}\right) d x = -\int \frac{1}{a+bu}\frac{1}{\sin^2x} dx = -\int \dfrac{1}{a+bu}\frac{-\sin^2 x}{\sin^2x} du. \end{align} Step 3: Evaluate the integral Now, evaluate the integral using the rule \(\int \dfrac{1}{k+lu} du =\dfrac{1}{l} \ln { |k+lu| }+C\) where C is the constant of integration: \begin{align} -\int \dfrac{1}{a+bu} du = -\ln{ |a+bu| }+C. \end{align} Step 4: Write the final answer Replace \(u\) with \(\cot x\): \begin{align} \int \dfrac{1}{a+b \cot x} d x = -\ln{ |a+b\cot x| }+C. \end{align}