Question

If $$\int {\frac{{\sin \,x}}{{\sin \left( {x - \alpha } \right)}}dx = Ax + B\log \,\sin \left( {x - \alpha } \right), + C,} $$         then value of $$\left( {A,\,B} \right)$$  is-

A. $$\left( { - \cos \,\alpha ,\,\sin \,\alpha } \right)$$
B. $$\left( {\cos \,\alpha ,\,\sin \,\alpha } \right)$$  
C. $$\left( { - \sin \,\alpha ,\,\cos \,\alpha } \right)$$
D. $$\left( {\sin \,\alpha ,\,\cos \,\alpha } \right)$$
Answer :   $$\left( {\cos \,\alpha ,\,\sin \,\alpha } \right)$$
Solution :
$$\eqalign{ & \int {\frac{{\sin \,x}}{{\sin \left( {x - \alpha } \right)}}dx} = \int {\frac{{\sin \,\left( {x - \alpha + \alpha } \right)}}{{\sin \left( {x - \alpha } \right)}}dx} \cr & = \int {\frac{{\sin \left( {x - \alpha } \right)\cos \,\alpha + \cos \left( {x - \alpha } \right)\sin \,\alpha }}{{\sin \left( {x - \alpha } \right)}}dx} \cr & = \int {\left\{ {\cos \,\alpha + \sin \,\alpha \,\cot \left( {x - \alpha } \right)} \right\}dx} \cr & = \left( {\cos \,\alpha } \right)x + \left( {\sin \,\alpha } \right)\log \,\sin \,\left( {x - \alpha } \right) + C \cr & \therefore A = \cos \,\alpha ,\,\,\,B = \sin \,\alpha \cr} $$

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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Indefinite Integration


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