the value of sin xsin 4x dx is 228028433

The value of $$\int {\frac{{\sin \,x}}{{\sin \,4x}}} dx$$ is :

A. $$\frac{1}{4}\log \left| {\frac{{\sin \,x - 1}}{{\sin \,x + 1}}} \right| - \frac{1}{{\sqrt 2 }}\log \left| {\frac{{\sqrt 2 \,\sin \,x - 1}}{{\sqrt 2 \,\sin \,x + 1}}} \right| + C$$

B. $$\frac{1}{8}\log \left| {\frac{{\cos \,x - 1}}{{\cos \,x + 1}}} \right| - \frac{1}{{2\sqrt 2 }}\log \left| {\frac{{\sqrt 2 \,\cos \,x - 1}}{{\sqrt 2 \,\cos \,x + 1}}} \right| + C$$

C. $$\frac{1}{8}\log \left| {\frac{{\sin \,x - 1}}{{\sin \,x + 1}}} \right| - \frac{1}{{4\sqrt 2 }}\log \left| {\frac{{\sqrt 2 \,\sin \,x - 1}}{{\sqrt 2 \,\sin \,x + 1}}} \right| + C$$

D. None of these

Answer : $$\frac{1}{8}\log \left| {\frac{{\sin \,x - 1}}{{\sin \,x + 1}}} \right| - \frac{1}{{4\sqrt 2 }}\log \left| {\frac{{\sqrt 2 \,\sin \,x - 1}}{{\sqrt 2 \,\sin \,x + 1}}} \right| + C$$

Solution :
$$\eqalign{ & I = \int {\frac{{\sin \,x\,dx}}{{4\,\sin \,x\,\cos \,x\,\cos \,2x}}} \cr & = \frac{1}{4}\int {\frac{{\cos \,x\,dx}}{{\left( {1 - {{\sin }^2}x} \right)\left( {1 - 2\,{{\sin }^2}x} \right)}}} \cr & = \frac{1}{4}\int {\frac{{dt}}{{\left( {1 - {t^2}} \right)\left( {1 - 2{t^2}} \right)}}\,\,\,\,\,\,\,\,\,\,\left[ {t = \sin \,x} \right]} \cr & = \frac{1}{4}\int {\left( {\frac{2}{{1 - 2{t^2}}} - \frac{1}{{1 - {t^2}}}} \right)dt} \cr & = \frac{1}{4}\left\{ {\frac{2}{{2\sqrt 2 }}\log \left| {\frac{{1 + \sqrt 2 t}}{{1 - \sqrt 2 t}}} \right| - \frac{1}{2}\log \left| {\frac{{1 + t}}{{1 - t}}} \right|} \right\} + C \cr & = \frac{1}{8}\log \left| {\frac{{\sin \,x - 1}}{{\sin \,x + 1}}} \right| - \frac{1}{{4\sqrt 2 }}\log \left| {\frac{{\sqrt 2 \,\sin \,x - 1}}{{\sqrt 2 \,\sin \,x + 1}}} \right| + C \cr} $$

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

View Answer

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. $$\sqrt 3 $$

View Answer

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. $$\frac{{\sqrt {2{x^4} - 2{x^2} + 1} }}{{2{x^2}}} + C$$

View Answer

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

View Answer

The value of $$\int {\frac{{\sin \,x}}{{\sin \,4x}}} dx$$ is :

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-

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

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

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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Indefinite Integration

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The value of $$\int {\frac{{\sin \,x}}{{\sin \,4x}}} dx$$ is :

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-

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

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

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-

Practice More Releted MCQ Question on Indefinite Integration

Practice More MCQ Question on Maths Section

Releted MCQ Question on
Calculus >> Indefinite Integration

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-

Practice More Releted MCQ Question on
Indefinite Integration