the integral sec 2 3 x cose c 4 3 x dx is equal to here c

The integral $$\int {{{\sec }^{\frac{2}{3}}}x\,{\text{cose}}{{\text{c}}^{\frac{4}{3}}}x\,dx} $$ is equal to:
(Here $$C$$ is a constant of integration)

A. $$ - 3\,{\tan ^{ - \,\frac{1}{3}}}x + C$$

B. $$ - \frac{3}{4}\,{\tan ^{ - \,\frac{4}{3}}}x + C$$

C. $$ - 3\,{\cot ^{ - \,\frac{1}{3}}}x + C$$

D. $$3\,{\tan ^{ - \,\frac{1}{3}}}x + C$$

Answer : $$ - 3\,{\tan ^{ - \,\frac{1}{3}}}x + C$$

Solution :
$$\eqalign{ & I = \int {\frac{{dx}}{{{{\left( {\sin \,x} \right)}^{\frac{4}{3}}}.{{\left( {\cos \,x} \right)}^{\frac{2}{3}}}}}} \cr & \Rightarrow I = \int {\frac{{dx}}{{{{\left( {\frac{{\sin \,x}}{{\cos \,x}}} \right)}^{\frac{4}{3}}}.{\cos^2}\,x}}} \cr & \Rightarrow I = \int {\frac{{{{\sec }^2}x}}{{{{\left( {\tan \,x} \right)}^{\frac{4}{3}}}}}dx} \cr & {\text{Put }}\tan \,x = t \Rightarrow {\sec ^2}x\,dx = dt \cr & \therefore \,I = \int {\frac{{dt}}{{{t^{\frac{4}{3}}}}}} \Rightarrow I = \frac{{ - 3}}{{{t^{\frac{1}{3}}}}} + C \cr & \Rightarrow I = \frac{{ - 3}}{{{{\left( {\tan \,x} \right)}^{\frac{1}{3}}}}} + C \cr & \Rightarrow I = - 3{\mkern 1mu} {\tan ^{ - {\kern 1pt} \frac{1}{3}}}x + C \cr} $$

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 integral $$\int {{{\sec }^{\frac{2}{3}}}x\,{\text{cose}}{{\text{c}}^{\frac{4}{3}}}x\,dx} $$ is equal to:
(Here $$C$$ is a constant of integration)

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

The integral $$\int {{{\sec }^{\frac{2}{3}}}x\,{\text{cose}}{{\text{c}}^{\frac{4}{3}}}x\,dx} $$ is equal to: (Here $$C$$ is a constant of integration)

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

The integral $$\int {{{\sec }^{\frac{2}{3}}}x\,{\text{cose}}{{\text{c}}^{\frac{4}{3}}}x\,dx} $$ is equal to:
(Here $$C$$ is a constant of integration)

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