हिंदी

Π / 2 ∫ 0 Cos 4 X D X - Mathematics

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
Definite Integrals
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 20: Definite Integrals - Exercise 20.1 [पृष्ठ १६]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 20 Definite Integrals
Exercise 20.1 | Q 22 | पृष्ठ १६

संबंधित प्रश्न

\[\int\limits_4^9 \frac{1}{\sqrt{x}} dx\]

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

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

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

\[\int\limits_1^e \frac{\log x}{x} dx\]

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

\[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\]

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

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

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

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

Evaluate the following integral:

\[\int\limits_{- 2}^2 \left| 2x + 3 \right| dx\]

\[\int\limits_0^\pi x \sin x \cos^4 x\ dx\]

\[\int\limits_0^3 \left( x + 4 \right) dx\]

\[\int\limits_{- 1}^1 \left( x + 3 \right) dx\]

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

\[\int\limits_{- 2}^1 \frac{\left| x \right|}{x} dx .\]

\[\int\limits_0^{\pi/2} \log \left( \frac{3 + 5 \cos x}{3 + 5 \sin x} \right) dx .\]

 


\[\int\limits_2^3 \frac{1}{x}dx\]

Evaluate each of the following integral:

\[\int_0^\frac{\pi}{4} \sin2xdx\]

Evaluate each of the following integral:

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

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

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

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

Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .


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


\[\int\limits_0^{2\pi} \cos^7 x dx\]


\[\int\limits_0^\pi x \sin x \cos^4 x dx\]


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


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


Prove that `int_a^b ƒ ("x") d"x" = int_a^bƒ(a + b - "x") d"x" and "hence evaluate" int_(π/6)^(π/3) (d"x")/(1+sqrt(tan "x")`


Using second fundamental theorem, evaluate the following:

`int_1^"e" ("d"x)/(x(1 + logx)^3`


Using second fundamental theorem, evaluate the following:

`int_(-1)^1 (2x + 3)/(x^2 + 3x + 7)  "d"x`


Evaluate the following:

`int_(-1)^1 "f"(x)  "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x  < 0):}`


Choose the correct alternative:

`int_0^oo x^4"e"^-x  "d"x` is


If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×