Question

The integral $$\int {\frac{{{{\sec }^2}x}}{{{{\left( {\sec \,x + \tan \,x} \right)}^{\frac{9}{2}}}}}dx} ,$$ equals (for some arbitrary constant $$K$$ )

A. $$ - \frac{1}{{{{\left( {\sec \,x + \tan \,x} \right)}^{\frac{{11}}{2}}}}}\left\{ {\frac{1}{{11}} - \frac{1}{7}{{\left( {\sec \,x + \tan \,x} \right)}^2}} \right\} + K$$

B. $$\frac{1}{{{{\left( {\sec \,x + \tan \,x} \right)}^{\frac{{11}}{2}}}}}\left\{ {\frac{1}{{11}} - \frac{1}{7}{{\left( {\sec \,x + \tan \,x} \right)}^2}} \right\} + K$$

C. $$ - \frac{1}{{{{\left( {\sec \,x + \tan \,x} \right)}^{\frac{{11}}{2}}}}}\left\{ {\frac{1}{{11}} + \frac{1}{7}{{\left( {\sec \,x + \tan \,x} \right)}^2}} \right\} + K$$

D. $$\frac{1}{{{{\left( {\sec \,x + \tan \,x} \right)}^{\frac{{11}}{2}}}}}\left\{ {\frac{1}{{11}} + \frac{1}{7}{{\left( {\sec \,x + \tan \,x} \right)}^2}} \right\} + K$$

Answer : $$ - \frac{1}{{{{\left( {\sec \,x + \tan \,x} \right)}^{\frac{{11}}{2}}}}}\left\{ {\frac{1}{{11}} + \frac{1}{7}{{\left( {\sec \,x + \tan \,x} \right)}^2}} \right\} + K$$

Solution :
$$\eqalign{ & I = \int {\frac{{{{\sec }^2}x}}{{{{\left( {\sec \,x + \tan \,x} \right)}^{\frac{9}{2}}}}}dx} \cr & {\text{Let }}\sec \,x + \tan \,x = t \cr & \Rightarrow \sec \,x - \tan \,x = \frac{1}{t} \cr & \Rightarrow \sec \,x = \frac{1}{2}\left( {t + \frac{1}{t}} \right) \cr & {\text{Also }}\sec x\left( {\sec \,x + \tan \,x} \right)dx = dt \cr & \Rightarrow \sec \,x\,dx = \frac{{dt}}{t} \cr & \therefore I = \frac{1}{2}\int {\frac{{\left( {t + \frac{1}{t}} \right)dt}}{{{t^{\frac{9}{2}}}.t}}} \cr & = \frac{1}{2}\int {\left( {{t^{ - \,\frac{9}{2}}} + {t^{ - \,\frac{{13}}{2}}}} \right)} dt \cr & = \frac{1}{2}\left[ {\frac{{{t^{ - \,\frac{9}{2}\, + \,1}}\,}}{{ - \frac{9}{2} + 1}} + \frac{{{t^{ - \,\frac{{13}}{2}\, + \,1}}}}{{ - \frac{{13}}{2} + 1}}} \right] + K \cr & = \frac{{ - 1}}{7}{t^{ - \,\frac{7}{2}}} - \frac{1}{{11}}{t^{ - \,\frac{{11}}{2}}} + K \cr & = - \frac{1}{{7{t^{\,\frac{7}{2}}}}} - \frac{1}{{11{t^{\,\frac{{11}}{2}}}}} + K \cr & = - \frac{1}{{{t^{\frac{{11}}{2}}}}}\left( {\frac{1}{{11}} + \frac{{{t^2}}}{7}} \right) + K \cr & = \frac{{ - 1}}{{{{\left( {\sec \,x + \tan \,x} \right)}^{\frac{{11}}{2}}}}}\left\{ {\frac{1}{{11}} + \frac{1}{7}{{\left( {\sec \,x + \tan \,x} \right)}^2}} \right\} + K \cr} $$

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. $$\sin \,x - 2{\left( {\sin \,x} \right)^{ - 1}} + 5\,{\tan ^{ - 1}}\left( {\sin \,x} \right) + c$$

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

The integral $$\int {\frac{{{{\sec }^2}x}}{{{{\left( {\sec \,x + \tan \,x} \right)}^{\frac{9}{2}}}}}dx} ,$$ equals (for some arbitrary constant $$K$$ )

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

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

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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-

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The integral $$\int {\frac{{{{\sec }^2}x}}{{{{\left( {\sec \,x + \tan \,x} \right)}^{\frac{9}{2}}}}}dx} ,$$ equals (for some arbitrary constant $$K$$ )

Releted MCQ Question on Calculus >> Indefinite Integration

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

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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-

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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-

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