मराठी

Π ∫ − π X 10 Sin 7 X D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]

बेरीज
Advertisements

उत्तर

\[\int_{- \pi}^\pi x^{10} \sin^7 x d x\]

\[Let f\left( x \right) = x^{10} \sin^7 x\]

\[\text{Consider }f\left( - x \right) = \left( - x \right)^{10} \sin^7 \left( - x \right) = - x^{10} \sin^7 x = - f\left( x \right)\]

Hence f(x) is an odd function

Therefore

\[ \int_{- \pi}^\pi x^{10} \sin^7 x d x = 0\]

shaalaa.com
Definite Integrals
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 20: Definite Integrals - Revision Exercise [पृष्ठ १२२]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 20 Definite Integrals
Revision Exercise | Q 55 | पृष्ठ १२२

संबंधित प्रश्‍न

\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]


\[\int\limits_{- \pi/4}^{\pi/4} \frac{1}{1 + \sin x} dx\]

\[\int\limits_0^{\pi/6} \cos x \cos 2x\ dx\]

\[\int\limits_0^{2\pi} e^{x/2} \sin\left( \frac{x}{2} + \frac{\pi}{4} \right) dx\]

\[\int\limits_0^1 \frac{1}{\sqrt{1 + x} - \sqrt{x}} dx\]

\[\int_0^\frac{\pi}{4} \left( a^2 \cos^2 x + b^2 \sin^2 x \right)dx\]

\[\int\limits_0^1 \frac{e^x}{1 + e^{2x}} dx\]

\[\int\limits_0^1 \frac{2x}{1 + x^4} dx\]

\[\int\limits_0^{\pi/2} \frac{dx}{a \cos x + b \sin x}a, b > 0\]

\[\int_0^\frac{\pi}{4} \frac{\sin x + \cos x}{3 + \sin2x}dx\]

\[\int\limits_1^2 \frac{1}{x \left( 1 + \log x \right)^2} dx\]

\[\int\limits_4^9 \frac{\sqrt{x}}{\left( 30 - x^{3/2} \right)^2} dx\]

\[\int\limits_0^1 \left( \cos^{- 1} x \right)^2 dx\]

\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{3/2}} dx\]

\[\int\limits_0^5 \frac{\sqrt[4]{x + 4}}{\sqrt[4]{x + 4} + \sqrt[4]{9 - x}} dx\]

\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]

If f (x) is a continuous function defined on [0, 2a]. Then, prove that

\[\int\limits_0^{2a} f\left( x \right) dx = \int\limits_0^a \left\{ f\left( x \right) + f\left( 2a - x \right) \right\} dx\]

 


\[\int\limits_1^2 x^2 dx\]

\[\int\limits_0^{\pi/2} \cos^2 x\ dx .\]

\[\int\limits_0^\infty e^{- x} dx .\]

If \[\int_0^a \frac{1}{4 + x^2}dx = \frac{\pi}{8}\] , find the value of a.


If \[\left[ \cdot \right] and \left\{ \cdot \right\}\] denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:

\[\int\limits_0^{\pi/4} \sin \left\{ x \right\} dx\]

 


\[\int\limits_0^{\pi/2} \frac{\cos x}{\left( 2 + \sin x \right)\left( 1 + \sin x \right)} dx\] equals

The value of \[\int\limits_0^{\pi/2} \cos x\ e^{\sin x}\ dx\] is

 


The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is

 


\[\int\limits_0^1 \frac{d}{dx}\left\{ \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \right\} dx\] is equal to

\[\int\limits_0^{\pi/2} x \sin x\ dx\]  is equal to

\[\int\limits_0^\infty \log\left( x + \frac{1}{x} \right) \frac{1}{1 + x^2} dx =\] 

\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]


\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]


\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]


Evaluate the following integrals as the limit of the sum:

`int_1^3 x  "d"x`


Choose the correct alternative:

`int_0^1 (2x + 1)  "d"x` is


Choose the correct alternative:

`int_0^oo "e"^(-2x)  "d"x` is


Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`


`int "e"^x ((1 - x)/(1 + x^2))^2  "d"x` is equal to ______.


Find: `int logx/(1 + log x)^2 dx`


The value of `int_2^3 x/(x^2 + 1)`dx is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×