what is pi 6 pi 6 sin 5xcos 3xx 4dx is equal to 215028539

What is $$\int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {\frac{{{{\sin }^5}x\,{{\cos }^3}x}}{{{x^4}}}dx} $$ is equal to ?

A. $$\frac{\pi }{2}$$

B. $$\frac{\pi }{4}$$

C. $$\frac{\pi }{8}$$

D. $$0$$

Answer : $$0$$

Solution :
$$\eqalign{ & {\text{Let }}f\left( x \right) = \frac{{{{\sin }^5}x\,{{\cos }^3}x}}{{{x^4}}} \cr & f\left( { - x} \right) = \frac{{{{\sin }^5}\left( { - x} \right)\,{{\cos }^3}\left( { - x} \right)}}{{{{\left( { - x} \right)}^4}}} \cr & f\left( { - x} \right) = \frac{{ - {{\sin }^5}x\,{{\cos }^3}x}}{{{x^4}}} \cr & f\left( { - x} \right) = - f\left( x \right) \cr} $$
$$ \Rightarrow f\left( x \right)$$ is an odd function.
Hence, $$\int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {\frac{{{{\sin }^5}x\,{{\cos }^3}x}}{{{x^4}}}dx} = 0$$

Releted MCQ Question on
Calculus >> Definite Integration

Releted Question 1

The value of the definite integral $$\int\limits_0^1 {\left( {1 + {e^{ - {x^2}}}} \right)} \,dx$$ is-

A. $$ - 1$$

B. $$2$$

C. $$1 + {e^{ - 1}}$$

D. none of these

View Answer

Releted Question 2

Let $$a,\,b,\,c$$ be non-zero real numbers such that $$\int\limits_0^1 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx = } \int\limits_0^2 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx.} $$
Then the quadratic equation $$a{x^2} + bx + c = 0$$ has-

A. no root in $$\left( {0,\,2} \right)$$

B. at least one root in $$\left( {0,\,2} \right)$$

C. a double root in $$\left( {0,\,2} \right)$$

D. two imaginary roots

View Answer

Releted Question 3

The value of the integral $$\int\limits_0^{\frac{\pi }{2}} {\frac{{\sqrt {\cot \,x} }}{{\sqrt {\cot \,x} + \sqrt {\tan \,x} }}dx} $$ is-

A. $$\frac{\pi }{4}$$

B. $$\frac{\pi }{2}$$

C. $$\pi $$

D. none of these

View Answer

Releted Question 4

For any integer $$n$$ the integral $$\int\limits_0^\pi {{e^{{{\cos }^2}x}}} {\cos ^3}\left( {2n + 1} \right)xdx$$ has the value-

A. $$\pi $$

B. $$1$$

C. $$0$$

D. none of these

View Answer

What is $$\int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {\frac{{{{\sin }^5}x\,{{\cos }^3}x}}{{{x^4}}}dx} $$ is equal to ?

Releted MCQ Question on
Calculus >> Definite Integration

The value of the definite integral $$\int\limits_0^1 {\left( {1 + {e^{ - {x^2}}}} \right)} \,dx$$ is-

Let $$a,\,b,\,c$$ be non-zero real numbers such that $$\int\limits_0^1 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx = } \int\limits_0^2 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx.} $$
Then the quadratic equation $$a{x^2} + bx + c = 0$$ has-

The value of the integral $$\int\limits_0^{\frac{\pi }{2}} {\frac{{\sqrt {\cot \,x} }}{{\sqrt {\cot \,x} + \sqrt {\tan \,x} }}dx} $$ is-

For any integer $$n$$ the integral $$\int\limits_0^\pi {{e^{{{\cos }^2}x}}} {\cos ^3}\left( {2n + 1} \right)xdx$$ has the value-

Practice More Releted MCQ Question on
Definite Integration

Practice More MCQ Question on Maths Section

What is $$\int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {\frac{{{{\sin }^5}x\,{{\cos }^3}x}}{{{x^4}}}dx} $$ is equal to ?

Releted MCQ Question on Calculus >> Definite Integration

The value of the definite integral $$\int\limits_0^1 {\left( {1 + {e^{ - {x^2}}}} \right)} \,dx$$ is-

Let $$a,\,b,\,c$$ be non-zero real numbers such that $$\int\limits_0^1 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx = } \int\limits_0^2 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx.} $$ Then the quadratic equation $$a{x^2} + bx + c = 0$$ has-

The value of the integral $$\int\limits_0^{\frac{\pi }{2}} {\frac{{\sqrt {\cot \,x} }}{{\sqrt {\cot \,x} + \sqrt {\tan \,x} }}dx} $$ is-

For any integer $$n$$ the integral $$\int\limits_0^\pi {{e^{{{\cos }^2}x}}} {\cos ^3}\left( {2n + 1} \right)xdx$$ has the value-

Practice More Releted MCQ Question on Definite Integration

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Releted MCQ Question on
Calculus >> Definite Integration

Let $$a,\,b,\,c$$ be non-zero real numbers such that $$\int\limits_0^1 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx = } \int\limits_0^2 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx.} $$
Then the quadratic equation $$a{x^2} + bx + c = 0$$ has-

Practice More Releted MCQ Question on
Definite Integration