Let $$\int {f\left( x \right)dx = \psi \left( x \right)} $$

Let $$I = \int {{x^5}} f\left( {{x^3}} \right)dx$$

put $${x^3} = t \Rightarrow 3{x^2}dx = dt$$

$$I = {1 \over 3}\int {3.{x^2}} .{x^3}.f\left( {{x^3}} \right).dx$$

$$ = {1 \over 3}\int {tf} \left( t \right)dt$$

$$ = {1 \over 3}\left[ {t\int {f\left( t \right)dt - \int {f\left( t \right)dt} } } \right]$$

$$ = {1 \over 3}\left[ {t\psi \left( t \right) - \int {\psi \left( t \right)dt} } \right]$$

$$ = {1 \over 3}\left[ {{x^3}\psi \left( {{x^3}} \right) - 3\int {{x^2}\psi \left( {{x^3}} \right)dx} } \right] + C$$

$$ = {1 \over 3}{x^3}\psi \left( {{x^3}} \right) - \int {{x^2}} \psi \left( {{x^3}} \right)dx + C$$

$$\int {{{5\tan x} \over {\tan x - 2}}} dx$$

$$ = \int {{{5{{\sin x} \over {\cos x}}} \over {{{\sin x} \over {\cos x}} - 2}}} \,dx$$

$$ = \int {\left( {{{5\sin x} \over {\cos x}} \times {{\cos x} \over {\sin x - 2\cos x}}} \right)} \,dx$$

$$ = \int {{{5\,\sin \,x\,dx} \over {\sin x - 2\,\cos x}}} $$

$$ = \int {\left( {{{4\sin x + \sin x + 2\cos x - 2\cos x} \over {\sin x - 2\cos x}}} \right)} \,dx$$

$$ = \int {{{\left( {\sin x - 2\cos x} \right) + \left( {4\sin x + 2\cos x} \right)} \over {\sin x - 2\cos x}}} \,dx$$

$$ = \int {{{\left( {\sin x - 2\cos x} \right) + 2\left( {\cos x + 2\sin x} \right)} \over {\left( {\sin x - 2\cos x} \right)}}} \,dx$$

$$ = \int {{{\sin x - 2\cos x} \over {\sin x - 2\cos x}}dx + 2\int {\left( {{{\cos x + 2\sin x} \over {\sin x - 2\cos x}}} \right)} } \,dx$$

$$ = \int {dx + 2\int {{{\cos x + 2\sin x} \over {\sin x - 2\cos \,x}}} } \,dx$$

$$ = {I_1} + {I_2}$$ where $${I_1} = \int {dx} $$ and

$${I_2} = 2\int {{{\cos x + 2\sin x} \over {\sin x - 2\cos x}}\,dx} $$

Put $$\sin x - 2\cos x = t$$

$$ \Rightarrow \left( {\cos x + 2\sin x} \right)dx = dt$$

$$\therefore$$ $${I_2} = 2\int {{{dt} \over t}} = 2\ln \,t + C$$

$$ = 2\,\ln \left( {\sin \,x - 2\cos \,x} \right) + C$$

Hence, $${I_1} + {I_2}$$

$$ = \int {dx + 2\ln \left( {\sin x - 2\cos x} \right) + c} $$

$$ = x + 2\ln \left| {\left( {\sin x - 2\cos x} \right)} \right| + k$$

$$ \Rightarrow a = 2$$

Let $$I = \sqrt 2 \int {{{\sin \,xdx} \over {\sin \left( {x - {\pi \over 4}} \right)}}} $$

Put $$x - {\pi \over 4} = t$$

$$ \Rightarrow dx = dt$$

$$ \Rightarrow I = \sqrt 2 \int {{{\sin \left( {t + {\pi \over 4}} \right)} \over {\sin \,t}}} dt$$

$$\,\,\,\,\,\,\,\,\,\,\,\, = {{\sqrt 2 } \over {\sqrt 2 }}\int {\left( {{{\sin t + \cos t} \over {\sin t}}} \right)} \,\,dt$$

$$ \Rightarrow I = \int {\left( {1 + \cot \,t} \right)} dt$$

$$\,\,\,\,\,\,\,\,\,\,\,\, = t + \log \left| {\sin t} \right| + {c_1}$$

$$ = x - {\pi \over 4} + \log \left| {\sin \left( {x - {\pi \over 4}} \right)} \right| + {c_1}$$

$$ = x + \log \left| {\sin \left( {x - {\pi \over 4}} \right)} \right| + c$$

$$\left( \, \right.$$ where $${c = {c_1} - {\pi \over 4}}$$ $$\left. \, \right)$$

$$I = \int {{{dx} \over {\cos x + \sqrt 3 \sin x}}} $$

$$ \Rightarrow I = \int {{{dx} \over {2\left[ {{1 \over 2}\cos x + {{\sqrt 3 } \over 2}\sin x} \right]}}} $$

$$ = {1 \over 2}\int {{{dx} \over {\left[ {\sin {\pi \over 6}\cos x + \cos {\pi \over 6}\sin x} \right]}}} $$

$$ = {1 \over 2}.\int {{{dx} \over {\sin \left( {x + {\pi \over 6}} \right)}}} $$

$$ \Rightarrow I = {1 \over 2}.\int {\cos ec\left( {x + {\pi \over 6}} \right)dx} $$

But we know that

$$\int {\cos ec\,x\,dx} = \log \left| {\left( {\tan x/2} \right)} \right| + C$$

$$\therefore$$ $$I = {1 \over 2}.\log \,\tan \left( {{x \over 2} + {\pi \over {12}}} \right) + C$$