हिंदी

Π / 2 ∫ − π / 2 Sin 3 X D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\limits_{- \pi/2}^{\pi/2} \sin^3 x\ dx\]

उत्तर

\[Let I = \int_\frac{- \pi}{2}^\frac{\pi}{2} \sin^3 x d x\]
\[ = \int_\frac{- \pi}{2}^\frac{\pi}{2} \sin x\left( 1 - \cos^2 x \right)dx\]
\[ = \int_\frac{- \pi}{2}^\frac{\pi}{2} \sin x dx - \int_\frac{- \pi}{2}^\frac{\pi}{2} \sin x \cos^2 x dx\]
\[ = \left[ - \cos x \right]_\frac{- \pi}{2}^\frac{\pi}{2} + \left[ \frac{\cos^3 x}{3} \right]_{- \frac{\pi}{2}}^\frac{\pi}{2} \]
\[ = 0 + 0 = 0\]

shaalaa.com
Definite Integrals
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 20: Definite Integrals - Exercise 20.5 [पृष्ठ ९५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 20 Definite Integrals
Exercise 20.5 | Q 23 | पृष्ठ ९५

संबंधित प्रश्न

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

\[\int\limits_0^{\pi/4} \sec x dx\]

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

\[\int\limits_{\pi/3}^{\pi/4} \left( \tan x + \cot x \right)^2 dx\]

\[\int\limits_0^{\pi/2} \sqrt{1 + \sin x}\ dx\]

Evaluate the following definite integrals:

\[\int_0^\frac{\pi}{2} x^2 \sin\ x\ dx\]

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

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

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

\[\int\limits_0^2 \frac{1}{\sqrt{3 + 2x - x^2}} dx\]

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

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

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

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

\[\int\limits_0^{\pi/2} \frac{x + \sin x}{1 + \cos x} dx\]

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

\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]


\[\int\limits_{- a}^a \sqrt{\frac{a - x}{a + x}} dx\]

\[\int_0^\pi \cos x\left| \cos x \right|dx\]

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

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

If f(x) is a continuous function defined on [−aa], then prove that 

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

\[\int\limits_3^5 \left( 2 - x \right) dx\]

\[\int\limits_1^2 \left( x^2 - 1 \right) dx\]

\[\int\limits_a^b x\ dx\]

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

\[\int\limits_0^\sqrt{2} \left[ x^2 \right] dx .\]

\[\int\limits_{\pi/6}^{\pi/3} \frac{1}{1 + \sqrt{\cot}x} dx\] is

\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan x} dx\]  is equal to

\[\int\limits_0^{2a} f\left( x \right) dx\]  is equal to


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


Using second fundamental theorem, evaluate the following:

`int_0^(pi/2) sqrt(1 + cos x)  "d"x`


Evaluate the following:

`int_0^2 "f"(x)  "d"x` where f(x) = `{{:(3 - 2x - x^2",", x ≤ 1),(x^2 + 2x - 3",", 1 < x ≤ 2):}`


Evaluate the following using properties of definite integral:

`int_0^(i/2) (sin^7x)/(sin^7x + cos^7x)  "d"x`


Choose the correct alternative:

`int_(-1)^1 x^3 "e"^(x^4)  "d"x` is


Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`


Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×