$$\int {\frac{{\frac{1}{2}d\left( {{x^2}} \right)}}{{1 + {{\left( {{x^2}} \right)}^2}}} = \frac{1}{2}{{\tan }^{ - 1}}\left( {{x^2}} \right) + k} $$

$$\eqalign{ & f'\left( x \right) = \int {{{\tan }^2}x\,dx = \int {\left( {{{\sec }^2}x - 1} \right)dx} = \tan \,x - x + k} \cr & \therefore f'\left( 0 \right) = k = 0 \cr & \therefore f'\left( x \right) = \tan \,x - x \cr & \therefore f\left( x \right) = \int {\left( {\tan \,x - x} \right)dx = \log \,\sec \,x - \frac{{{x^2}}}{2} + c} \cr & \therefore f\left( 0 \right) = c = 0 \cr} $$

$$f\left( x \right) = x\left| {\cos \,x} \right|,\,\frac{\pi }{2} < x < \pi $$
$$= - x\cos \,x,$$    because $$\cos \,x$$   is negative in $$\left( {\frac{\pi }{2},\,\pi } \right)$$
$$\therefore $$ the required primitive function $$ = \int { - x\cos \,x\,dx} $$
Now use integration by parts.

Here $$\int {{e^x}\left\{ {f\left( x \right) - f'\left( x \right)} \right\}dx = \phi \left( x \right)} $$       and $$\int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}dx = {e^x}f\left( x \right)} .$$
Adding, $$2\int {{e^x}f\left( x \right)dx = \phi \left( x \right) + {e^x}f\left( x \right)} $$

$$\eqalign{ & I = \int {{{\sin }^3}x.{{\cos }^5}x} \,dx \cr & {\text{Put }}\sin \,x = t \Rightarrow \cos \,x\,dx = dt \cr & I = \int {{{\sin }^3}x.{{\cos }^4}x} .\cos \,x\,dx \cr & \,\,\,\,\, = \int {{t^3}{{\left( {1 - {t^2}} \right)}^2}dt} \cr & \,\,\,\,\, = \int {\left( {{t^3} - 2{t^5} + {t^7}} \right)} dt \cr & \,\,\,\,\, = \frac{1}{4}{t^4} - \frac{2}{6}{t^6} + \frac{1}{8}{t^8} + D \cr & \,\,\,\,\, = \frac{1}{4}{\sin ^4}x - \frac{1}{3}{\sin ^6}x + \frac{1}{8}{\sin ^8}x + D \cr} $$

$$\eqalign{ & \int {{x^{51}}\left( {{{\tan }^{ - 1}}x + {{\cot }^{ - 1}}x} \right)dx} \cr & = \int {{x^{51}}.\frac{\pi }{2}dx} \,\,\,\,\,\left\{ {\therefore \,{{\tan }^{ - 1}}x + {{\cot }^{ - 1}}x = \frac{\pi }{2}} \right\} \cr & = \frac{{\pi {x^{52}}}}{{104}} + c \cr & = \frac{{{x^{52}}}}{{52}}\left( {{{\tan }^{ - 1}}x + {{\cot }^{ - 1}}x} \right) + c \cr} $$

$$\eqalign{ & \int {\frac{{{{\left( {1 + x} \right)}^2}}}{{x + {x^3}}}dx} = \int {\frac{{\left( {1 + {x^2}} \right) + 2x}}{{x\left( {1 + {x^2}} \right)}}dx} \cr & = \int {\left( {\frac{1}{x} + \frac{2}{{1 + {x^2}}}} \right)} dx \cr & = \log \,x + 2{\tan ^{ - 1}}x + k \cr} $$

$$\eqalign{ & {\text{Let,}} \cr & I = \int {\frac{1}{{1 + \sin \,x}}dx} \cr & \,\,\,\,\, = \int {\frac{{dx}}{{1 + \frac{{2\,\tan \frac{x}{2}}}{{1 + {{\tan }^2}\frac{x}{2}}}}}} \cr & \,\,\,\,\, = \int {\frac{{\left( {1 + {{\tan }^2}\frac{x}{2}} \right)dx}}{{1 + {{\tan }^2}\frac{x}{2} + 2\,\tan \,\frac{x}{2}}}} \cr & \,\,\,\,\, = \int {\frac{{{{\sec }^2}\frac{x}{2}dx}}{{1 + {{\tan }^2}\frac{x}{2} + 2\,\tan \,\frac{x}{2}}}} \cr & {\text{Substitute,}}\, \cr & \tan \frac{x}{2} = t \cr & \Rightarrow \frac{1}{2}{\sec ^2}\frac{x}{2}dx = dt \cr & \Rightarrow {\sec ^2}\frac{x}{2}dx = 2\,dt \cr & {\text{Then}} \cr & I = \int {\frac{{2dt}}{{1 + {t^2} + 2t}}} \cr & \,\,\,\,\, = 2\int {\frac{{dt}}{{{{\left( {1 + t} \right)}^2}}}} \cr & \,\,\,\,\, = 2\frac{{ - 1}}{{\left( {1 + t} \right)}} + C \cr & \,\,\,\,\, = \frac{{ - 2}}{{1 + \tan \frac{x}{2}}} + c \cr & \,\,\,\,\, = 1 - \frac{2}{{1 + \tan \frac{x}{2}}} + \left( {c - 1} \right) \cr & \,\,\,\,\, = \frac{{\tan \frac{x}{2} - 1}}{{\tan \frac{x}{2} + 1}} + b \cr & {\text{Where }}b = c - 1,{\text{ a new constant}} \cr & \,\,\,\,\, = - \frac{{1 - \tan \frac{x}{2}}}{{1 + \tan \frac{x}{2}}} + b \cr & \,\,\,\,\, = - \tan \left( {\frac{\pi }{4} - \frac{x}{2}} \right) + b \cr & \,\,\,\,\, = \tan \left( {\frac{x}{2} - \frac{\pi }{4}} \right) + b \cr & {\text{Clearly,}}\,a = - \frac{\pi }{4}{\text{ and }}b\, \in {\bf{R}} \cr} $$

$$\eqalign{ & {\text{Let }}I = \int {\frac{{\left( {1 + x} \right){e^x}}}{{\cot \left( {x{e^x}} \right)}}dx} \cr & {\text{Put }}x{e^x} = t \cr & \Rightarrow \left( {x{e^x} + {e^x}} \right)dx = dt \cr & \Rightarrow {e^x}\left( {x + 1} \right)dx = dt \cr & \therefore \,I = \int {\frac{{dt}}{{\cot \left( t \right)}}} \cr & = \log \left| {\sec \,t} \right| + C \cr & = \log \left| {\sec \left( {x{e^x}} \right)} \right| + C \cr} $$

Given that
$$\eqalign{ & \int_{\sin \,x}^1 {{t^2}f\left( t \right)dt = 1 - \sin \,x} \cr & \Rightarrow \frac{d}{{dx}}\int_{\sin \,x}^1 {{t^2}f\left( t \right)dt = \frac{d}{{dx}}\left( {1 - \sin \,x} \right)} \cr & \Rightarrow - {\sin ^2}xf\left( {\sin \,x} \right).cos\,x = - \cos \,x \cr & \Rightarrow f\left( {\sin \,x} \right) = \frac{1}{{{{\sin }^2}\,x}} \cr & \Rightarrow f\left( {\frac{1}{{\sqrt 3 }}} \right) = \frac{1}{{{{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}}} = 3 \cr} $$