ক) সমীকরণ কর:- 

  $\int \frac{\sin{x} + \cos{2} x}{1 - \sin{x}} dx$  

খ) মান নির্ণয় করঃ

  $\int_{0}^{\frac{π}{2}} \frac{dx}{\sin{x} + \cos{x}}$  

ক)

ত্রিকোণমিতিক অভেদ ব্যবহার কর্রে $\cos 2 x$ কে প্রকাশ কর্রা:

$\cos 2 x=1-2 \sin ^{2} x$

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

$\frac{\sin x+1-2 \sin ^{2} x}{1-\sin x}=\frac{1+\sin x-2 \sin ^{2} x}{1-\sin x}$

$1+\sin x-2 \sin ^{2} x=-2 \sin ^{2} x+\sin x+1$$\frac{-2 \sin ^{2} x+\sin x+1}{1-\sin x}=\frac{-\left(2 \sin ^{2} x-\sin x-1\right)}{1-\sin x}=\frac{2 \sin ^{2} x-\sin x-1}{\sin x-1}$

$2 \sin ^{2} x-\sin x-1=(2 \sin x+1)(\sin x-1)$

$\left.\frac{(2 \sin x+1)(\sin x-1)}{\sin x-1}=2 \sin x+1 \quad \text { (যখন } \sin x \neq 1\right)$

$\int(2 \sin x+1) d x=2 \int \sin x d x+\int 1 d x=-2 \cos x+x+C$

$\begin{array}{l} \text { খ) } \int_{0}^{\pi / 2} \frac{d x}{\sin x+\cos x}=\int_{0}^{\pi / 2} \frac{d x}{2 \sin \frac{x}{2} \cos \frac{x}{2}+\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}} \\ \int_{0}^{\pi / 2} \frac{\sec ^{2} \frac{x}{2} d x}{2 \tan \frac{x}{2}+1-\tan ^{2} \frac{x}{2}} \therefore \int_{0}^{1} \frac{2 d z}{2 z+1-z^{2}} \end{array}$

[Let, $\tan \frac{x}{2}=\mathrm{z} \Rightarrow \sec ^{2} \frac{\mathrm{x}}{2} \mathrm{dx}=2 \mathrm{dz} ; \mathrm{x}=\frac{\pi}{2}$ হলে $\mathrm{z}=1 ; \mathrm{x}=0$ হলে $\mathrm{z}=0$ ]

$\begin{array}{l} \therefore \mathrm{I}=2 \int_{0}^{1} \frac{\mathrm{dz}}{(\sqrt{2})^{2}-(\mathrm{z}-1)^{2}}=2 \frac{1}{2 \sqrt{2}}\left[\log \frac{\sqrt{2}+z-1}{\sqrt{2}-z+1}\right]_{0}^{1} \\ =\frac{1}{\sqrt{2}}\left(\log \frac{\sqrt{2}}{\sqrt{2}}-\log \frac{\sqrt{2}-1}{\sqrt{2}+1}\right)=\frac{1}{\sqrt{2}} \log \frac{\sqrt{2}+1}{\sqrt{2}-1} \end{array}$