81.
$$\int {\sqrt {\frac{x}{{1 - x}}} } dx$$ is equal to :
A
$${\sin ^{ - 1}}\sqrt x + c$$
B
$${\sin ^{ - 1}}\left\{ {\sqrt x - \sqrt {x\left( {1 - x} \right)} } \right\} + c$$
C
$${\sin ^{ - 1}}\sqrt {x\left( {1 - x} \right)} + c$$
D
$${\sin ^{ - 1}}\sqrt x - \sqrt {x\left( {1 - x} \right)} + c$$
Answer :
$${\sin ^{ - 1}}\sqrt x - \sqrt {x\left( {1 - x} \right)} + c$$
$$\eqalign{
& {\text{Put }}x = {\sin ^2}\theta \Rightarrow dx = 2\,\sin \,\theta \,\cos \,\theta \cr
& \therefore \,\int {\sqrt {\frac{x}{{1 - x}}} dx} \cr
& = \int {\frac{{\sin \,\theta }}{{\cos \,\theta }}.2\,\sin \,\theta \,\cos \,\theta } \,d\theta \cr
& = \int {\left( {1 - \cos \,2\theta } \right)} d\theta \cr
& = \theta - \frac{1}{2}\sin \,2\theta + c \cr
& = {\sin ^{ - 1}}\sqrt x - \sqrt {x\left( {1 - x} \right)} + c \cr} $$
82.
$$\int {\sin \,2x} .\log \,\cos \,x\,dx$$ is equal to :
A
$${\cos ^2}x\left( {\frac{1}{2} + \log \,\cos \,x} \right) + k$$
B
$${\cos ^2}x.\log \,\cos \,x + k$$
C
$${\cos ^2}x\left( {\frac{1}{2} - \log \,\cos \,x} \right) + k$$
D
none of these
Answer :
$${\cos ^2}x\left( {\frac{1}{2} - \log \,\cos \,x} \right) + k$$
$$\eqalign{
& I = \int {2\sin \,x.\cos \,x.\log \,\cos \,x\,dx} \cr
& {\text{Put }}\log {\text{ }}\cos {\text{ }}x = z \cr
& \therefore - \frac{{\sin \,x}}{{\cos \,x}}dx = dz \cr
& \therefore I = \int {2\sin \,x.\cos \,x.z.\frac{{\cos \,x}}{{ - \sin \,x}}dz} \cr
& = - 2\int {{{\cos }^2}x.z\,dz} \cr
& = - 2\int {z{e^{2z}}dz} \cr
& = - 2\left[ {z.\frac{{{e^{2z}}}}{2} - \int {\frac{{{e^{2z}}}}{2}.dz} } \right] \cr
& = - z.{e^{2z}} + \frac{1}{2}{e^{2z}} + k \cr
& = {e^{2z}}\left( {\frac{1}{2} - z} \right) + k \cr
& = {\cos ^2}x.\left\{ {\frac{1}{2} - \log \,\cos \,x} \right\} + k \cr} $$