हिंदी

Π / 2 ∫ 0 Cos 4 X D X

Advertisements
Advertisements

प्रश्न

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

 

योग
Advertisements

उत्तर

\[Let\ I = \int_0^\frac{\pi}{2} \cos^4 x\ dx\ . Then, \]
\[I = \int_0^\frac{\pi}{2} \left( \cos^2 x \right)^2 dx\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} \frac{\left( 1 + \cos 2x \right)^2}{4} dx\]
\[ \Rightarrow I = \frac{1}{4} \int_0^\frac{\pi}{2} \left( 1 + \cos^2 2x + 2 \cos 2x \right) dx\]
\[ \Rightarrow I = \frac{1}{4} \int_0^\frac{\pi}{2} \left( 1 + 2 \cos 2x + \frac{1 + \cos 4x}{2} \right) dx\]
\[ \Rightarrow I = \frac{1}{4} \int_0^\frac{\pi}{2} \left( \frac{3 + 4 \cos 2x + \cos 4x}{2} \right) dx\]
\[ \Rightarrow I = \frac{1}{4} \left[ \frac{3x}{2} + \frac{2 \sin 2x}{2} + \frac{\sin 4x}{8} \right]_0^\frac{\pi}{2} \]
\[ \Rightarrow I = \frac{1}{4}\left[ \frac{3\pi}{4} + 0 \right]\]
\[ \Rightarrow I = \frac{3\pi}{16}\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Definite Integrals - Exercise 20.1 [पृष्ठ १६]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 19 Definite Integrals
Exercise 20.1 | Q 22 | पृष्ठ १६

संबंधित प्रश्न

\[\int\limits_{- 2}^3 \frac{1}{x + 7} dx\]

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

\[\int\limits_{- 1}^1 \frac{1}{1 + x^2} dx\]

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

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

\[\int\limits_0^a \frac{x}{\sqrt{a^2 + x^2}} dx\]

\[\int\limits_0^{\pi/2} \frac{\sin \theta}{\sqrt{1 + \cos \theta}} d\theta\]

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

\[\int\limits_{- 1}^1 5 x^4 \sqrt{x^5 + 1} dx\]

\[\int_0^\frac{1}{2} \frac{1}{\left( 1 + x^2 \right)\sqrt{1 - x^2}}dx\]

\[\int_0^1 | x\sin \pi x | dx\]

\[\int\limits_0^2 \left( x + 3 \right) dx\]

\[\int\limits_1^3 \left( 3x - 2 \right) dx\]

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

\[\int\limits_2^3 \left( 2 x^2 + 1 \right) dx\]

\[\int\limits_0^5 \left( x + 1 \right) dx\]

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

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

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

Evaluate each of the following integral:

\[\int_0^\frac{\pi}{4} \tan\ xdx\]

 


Evaluate each of the following integral:

\[\int_e^{e^2} \frac{1}{x\log x}dx\]

The value of the integral \[\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\]  is 


`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`


\[\int\limits_{- \pi/2}^{\pi/2} \sin\left| x \right| dx\]  is equal to

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

 


\[\int\limits_{\pi/6}^{\pi/3} \frac{1}{\sin 2x} dx\]  is equal to

The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is 


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


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


\[\int\limits_0^\pi \cos 2x \log \sin x dx\]


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


\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]


\[\int\limits_0^4 x dx\]


\[\int\limits_0^2 \left( 2 x^2 + 3 \right) dx\]


\[\int\limits_{- 1}^1 e^{2x} dx\]


Evaluate the following using properties of definite integral:

`int_0^1 x/((1 - x)^(3/4))  "d"x`


If f(x) = `{{:(x^2"e"^(-2x)",", x ≥ 0),(0",", "otherwise"):}`, then evaluate `int_0^oo "f"(x) "d"x`


Evaluate `int (x^2 + x)/(x^4 - 9) "d"x`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×