Question
If $$\int {\sec \,x\,{\text{cosec}}\,x\,dx} = \log \left| {g\left( x \right)} \right| + c,$$ then what is $$g\left( x \right)$$ equal to ?
A.
$$\sin \,x\,\cos \,x$$
B.
$${\sec ^2}x$$
C.
$$\tan \,x$$
D.
$$\log \left| {\tan \,x} \right|$$
Answer :
$$\tan \,x$$
Solution :
$$\eqalign{
& {\text{Let }}I = \int {\sec \,x.{\text{cosec}}\,x\,dx} \cr
& = \int {\frac{2}{{2\,\sin \,x\,\cos \,x}}dx} \cr
& = 2\int {\frac{1}{{\sin \,2x}}dx - 2\int {\frac{1}{{\frac{{2\,\tan \,x}}{{1 + {{\tan }^2}x}}}}} } \,\,\,\,\,\left[ {\because \,\sin \,2x = \frac{{2\,\tan \,x}}{{1 + {{\tan }^2}x}}} \right] \cr
& = \int {\frac{{{{\sec }^2}x}}{{\tan \,x}}dx} \cr
& {\text{Let }}\tan \,x = t \Rightarrow {\sec ^2}dx = dt \cr
& {\text{So, }}I = \int {\frac{{dt}}{t}} = \log \left| t \right| + c = \log \left| {\tan \,x} \right| + c \cr
& {\text{But}}\,\int {\sec \,x\,{\text{cosec}}\,x\,dx = \log \left| {g\left( x \right)} \right|} + c \cr
& \therefore \,g\left( x \right) = \tan \,x \cr} $$