if i sin 11 3xcos 1 3xdx acot 2 3x bcot 8 3x c then

If $$I = \int {{{\sin }^{ - \frac{{11}}{3}}}x\,{{\cos }^{ - \frac{1}{3}}}x\,dx} = A\,{\cot ^{\frac{2}{3}}}x + B\,{\cot ^{\frac{8}{3}}}x + C.$$ Then :

A. $$A = \frac{2}{3},\,B = \frac{8}{3}$$

B. $$A = - \frac{3}{2},\,B = - \frac{3}{8}$$

C. $$A = \frac{3}{2},\,B = \frac{3}{8}$$

D. None of these

Answer : $$A = - \frac{3}{2},\,B = - \frac{3}{8}$$

Solution :
$$\eqalign{ & {\text{If,}} \cr & I = \int {{{\sin }^{ - \frac{{11}}{3}}}x\,{{\cos }^{ - \frac{1}{3}}}x\,dx} , \cr & {\text{here }}\frac{{ - \frac{{11}}{3} - \frac{1}{3} - 2}}{2} = - 3\,\,\left( {{\text{a negative integer}}} \right) \cr & I = \int {\frac{{{{\sin }^{ - \frac{{11}}{3}}}x}}{{{{\cos }^{ - \frac{{11}}{3}}}x}}} {\cos ^{ - \frac{1}{3}}}x.{\cos ^{ - \frac{{11}}{3}}}x\,dx \cr & I = \int {{{\left( {\tan \,x} \right)}^{ - \frac{{11}}{3}}}x.{{\left( {\cos \,x} \right)}^{ - 4}}dx} \cr & I = \int {{{\left( {\tan \,x} \right)}^{ - \frac{{11}}{3}}}x.{{\sec }^4}xdx} \cr & I = \int {{{\left( {\tan \,x} \right)}^{ - \frac{{11}}{3}}}\left( {1 + {{\tan }^2}\,x} \right){{\sec }^2}x\,dx} \cr & {\text{Put }}\tan \,x = t,\,{\sec ^2}x\,dx = dt \cr & I = \int {{t^{ - \frac{{11}}{3}}}} \left( {1 + {t^2}} \right)dt \cr & I = \int {\left( {{t^{ - \frac{{11}}{3}}} + {t^{ - \frac{5}{3}}}} \right)} dt \cr & I = \frac{{{t^{ - \frac{8}{3}}}}}{{ - \frac{8}{3}}} + \frac{{{t^{ - \frac{2}{3}}}}}{{ - \frac{2}{3}}} + C \cr & I = - \frac{3}{8}{\left( {\tan \,x} \right)^{ - \frac{8}{3}}} - \frac{3}{2}{\left( {\tan \,x} \right)^{ - \frac{2}{3}}} + C \cr & I = - \frac{3}{2}{\cot ^{\frac{2}{3}}}x - \frac{3}{8}{\cot ^{\frac{8}{3}}}x + C \cr} $$

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

If $$I = \int {{{\sin }^{ - \frac{{11}}{3}}}x\,{{\cos }^{ - \frac{1}{3}}}x\,dx} = A\,{\cot ^{\frac{2}{3}}}x + B\,{\cot ^{\frac{8}{3}}}x + C.$$ Then :

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

If $$I = \int {{{\sin }^{ - \frac{{11}}{3}}}x\,{{\cos }^{ - \frac{1}{3}}}x\,dx} = A\,{\cot ^{\frac{2}{3}}}x + B\,{\cot ^{\frac{8}{3}}}x + C.$$ Then :

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