int dx cos x 3 sin x equals 720001539

$$\int {\frac{{dx}}{{\cos \,x + \sqrt 3 \,\sin \,x}}} $$ equals-

A. $$\log \,\tan \,\left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) + C$$

B. $$\log \,\tan \,\left( {\frac{x}{2} - \frac{\pi }{{12}}} \right) + C$$

C. $$\frac{1}{2}\log \,\tan \,\left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) + C$$

D. $$\frac{1}{2}\log \,\tan \,\left( {\frac{x}{2} - \frac{\pi }{{12}}} \right) + C$$

Answer : $$\frac{1}{2}\log \,\tan \,\left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) + C$$

Solution :
$$\eqalign{ & I = \int {\frac{{dx}}{{\cos \,x + \sqrt 3 \,\sin \,x}}} \cr & \Rightarrow I = \int {\frac{{dx}}{{2\left[ {\frac{1}{2}\cos \,x + \frac{{\sqrt 3 }}{2}\sin \,x} \right]}}} \cr & \Rightarrow I = \frac{1}{2}\int {\frac{{dx}}{{\left[ {\sin \frac{\pi }{6}\cos \,x + \cos \frac{\pi }{6}\sin \,x} \right]}}} \cr & \Rightarrow I = \frac{1}{2}\int {\frac{{dx}}{{\sin \left( {x + \frac{\pi }{6}} \right)}}} \cr & \Rightarrow I = \frac{1}{2}\int {{\text{cosec}}\,\left( {x + \frac{\pi }{6}} \right)dx} \cr} $$
But we know that $$\int {{\text{cosec}}\,x\,dx} = \log \left| {\left( {\frac{{\tan \,x}}{2}} \right)} \right| + C$$
$$\therefore I = \frac{1}{2}.\log \,\tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) + C$$

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

$$\int {\frac{{dx}}{{\cos \,x + \sqrt 3 \,\sin \,x}}} $$ equals-

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

$$\int {\frac{{dx}}{{\cos \,x + \sqrt 3 \,\sin \,x}}} $$ equals-

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