যোগজীকরণ নির্ণয় কর:  

  $\int \frac{dx}{\cos{x} + \sin{x}}$  

• 
  

$\int \frac{d x}{\cos x+\sin x}=\int \frac{d x}{\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}+\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}}$

$\begin{array}{l} =\int \frac{\sec ^{2} \frac{x}{2} d x}{\left(-\tan ^{2} \frac{x}{2}+2 \tan \frac{x}{2}+1\right)} d x \\ =\int \frac{2 d x}{\left(-z^{2}+2 z+1\right)} \quad \text { Let, } \tan \frac{x}{2}=z \\ \Rightarrow 2 d z=\sec ^{2} \frac{x}{2} d x \end{array}$

$\begin{array}{l} =-2 \int \frac{\mathrm{dz}}{\left(\mathrm{z}^{2}-2 \mathrm{z}+1\right)-2} \\ =-2 \int \frac{\mathrm{dz}}{(\mathrm{z}-1)^{2}-(\sqrt{2})^{2}} \\ =-2 \frac{1}{2 \cdot \sqrt{2}} \ln \left|\frac{\mathrm{z}-1-\sqrt{2}}{\mathrm{z}-1+\sqrt{2}}\right| \\ =\frac{-1}{\sqrt{2}} \ln \left|\frac{\mathrm{z}-\sqrt{2}-1}{\mathrm{z}+\sqrt{2}-1}\right|+C=-\frac{1}{\sqrt{2}} \ln \left|\frac{\tan \frac{x}{2}-\sqrt{2}-1}{\tan \frac{x}{2}+\sqrt{2}-1}\right|+C \end{array}$