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 :
$$\eqalign{ & {\text{Let }}I = \int {\sec \,x.{\text{cosec}}\,x\,dx} \cr & = \int {\frac{2}{{2\,\sin \,x\,\cos \,x}}dx} \cr & = 2\int {\frac{1}{{\sin \,2x}}dx - 2\int {\frac{1}{{\frac{{2\,\tan \,x}}{{1 + {{\tan }^2}x}}}}} } \,\,\,\,\,\left[ {\because \,\sin \,2x = \frac{{2\,\tan \,x}}{{1 + {{\tan }^2}x}}} \right] \cr & = \int {\frac{{{{\sec }^2}x}}{{\tan \,x}}dx} \cr & {\text{Let }}\tan \,x = t \Rightarrow {\sec ^2}dx = dt \cr & {\text{So, }}I = \int {\frac{{dt}}{t}} = \log \left| t \right| + c = \log \left| {\tan \,x} \right| + c \cr & {\text{But}}\,\int {\sec \,x\,{\text{cosec}}\,x\,dx = \log \left| {g\left( x \right)} \right|} + c \cr & \therefore \,g\left( x \right) = \tan \,x \cr} $$

Releted MCQ Question on
Calculus >> Indefinite Integration

Releted Question 1

The value of the integral $$\int {\frac{{{{\cos }^3}x + {{\cos }^5}x}}{{{{\sin }^2}x + {{\sin }^4}x}}dx} $$    is-

A. $$\sin \,x - 6\,{\tan ^{ - 1}}\left( {\sin \,x} \right) + c$$
B. $$\sin \,x - 2{\left( {\sin \,x} \right)^{ - 1}} + c$$
C. $$\sin \,x - 2{\left( {\sin \,x} \right)^{ - 1}} - 6\,{\tan ^{ - 1}}\left( {\sin \,x} \right) + c$$
D. $$\sin \,x - 2{\left( {\sin \,x} \right)^{ - 1}} + 5\,{\tan ^{ - 1}}\left( {\sin \,x} \right) + c$$
Releted Question 2

If $$\int_{\sin \,x}^1 {{t^2}f\left( t \right)dt = 1 - \sin \,x} ,$$      then $$f\left( {\frac{1}{{\sqrt 3 }}} \right)$$   is-

A. $$\frac{1}{3}$$
B. $${\frac{1}{{\sqrt 3 }}}$$
C. $$3$$
D. $$\sqrt 3 $$
Releted Question 3

Solve this $$\int {\frac{{{x^2} - 1}}{{{x^3}\sqrt {2{x^4} - 2{x^2} + 1} }}dx} = ?$$

A. $$\frac{{\sqrt {2{x^4} - 2{x^2} + 1} }}{{{x^2}}} + C$$
B. $$\frac{{\sqrt {2{x^4} - 2{x^2} + 1} }}{{{x^3}}} + C$$
C. $$\frac{{\sqrt {2{x^4} - 2{x^2} + 1} }}{x} + C$$
D. $$\frac{{\sqrt {2{x^4} - 2{x^2} + 1} }}{{2{x^2}}} + C$$
Releted Question 4

Let $$I = \int {\frac{{{e^x}}}{{{e^{4x}} + {e^{2x}} + 1}}dx,\,J = \int {\frac{{{e^{ - \,x}}}}{{{e^{ - \,4x}} + {e^{ - \,2x}} + 1}}dx.} } $$
Then for an arbitrary constant $$C,$$ the value of $$J-I$$  equals-

A. $$\frac{1}{2}\log \left( {\frac{{{e^{4x}} - {e^{2x}} + 1}}{{{e^{4x}} + {e^{2x}} + 1}}} \right) + C$$
B. $$\frac{1}{2}\log \left( {\frac{{{e^{2x}} + {e^x} + 1}}{{{e^{2x}} - {e^x} + 1}}} \right) + C$$
C. $$\frac{1}{2}\log \left( {\frac{{{e^{2x}} - {e^x} + 1}}{{{e^{2x}} + {e^x} + 1}}} \right) + C$$
D. $$\frac{1}{2}\log \left( {\frac{{{e^{4x}} + {e^{2x}} + 1}}{{{e^{4x}} - {e^{2x}} + 1}}} \right) + C$$

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