मराठी

Π / 2 ∫ 0 Cos 3 X D X

Advertisements
Advertisements

प्रश्न

\[\int\limits_0^{\pi/2} \cos^3 x\ dx\]
Advertisements

उत्तर

\[Let\ I = \int_0^\frac{\pi}{2} \cos^3 x\ d\ x . Then, \]
\[I = \int_0^\frac{\pi}{2} \cos^2 x \cos\ x\ d\ x\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} \left( 1 - \sin^2 x \right) \cos x d x\]
\[Let u = \sin x, du = \cos\ x\ dx\]
\[ \Rightarrow I = \int\left( 1 - u^2 \right) du\]
\[ \Rightarrow I = \left[ u - \frac{u^3}{3} \right]\]
\[ \Rightarrow I = \left[ \sin x - \frac{\sin^3 x}{3} \right]_0^\frac{\pi}{2} \]
\[ \Rightarrow I = 1 - \frac{1}{3} - 0\]
\[ \Rightarrow I = \frac{2}{3}\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Definite Integrals - Exercise 20.1 [पृष्ठ १६]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
Exercise 20.1 | Q 18 | पृष्ठ १६

संबंधित प्रश्‍न

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

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

\[\int\limits_1^2 \frac{3x}{9 x^2 - 1} dx\]

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

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

\[\int\limits_0^{\pi/4} \frac{\tan^3 x}{1 + \cos 2x} dx\]

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

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

\[\int_0^\frac{\pi}{2} \sqrt{\cos x - \cos^3 x}\left( \sec^2 x - 1 \right) \cos^2 xdx\]

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

\[\int_{- \frac{\pi}{4}}^\frac{\pi}{2} \sin x\left| \sin x \right|dx\]

 


\[\int\limits_0^7 \frac{\sqrt[3]{x}}{\sqrt[3]{x} + \sqrt[3]{7} - x} dx\]

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

\[\int\limits_0^\pi x \sin^3 x\ dx\]

\[\int\limits_1^4 \left( x^2 - x \right) dx\]

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

\[\int\limits_a^b \frac{f\left( x \right)}{f\left( x \right) + f\left( a + b - x \right)} dx .\]

Evaluate : 

\[\int\limits_2^3 3^x dx .\]

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

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

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

Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .


\[\int\limits_1^2 x\sqrt{3x - 2} dx\]


\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]


Evaluate the following integrals :-

\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]


\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]


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


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


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


Find : `∫_a^b logx/x` dx


Using second fundamental theorem, evaluate the following:

`int_0^1 x"e"^(x^2)  "d"x`


Evaluate the following:

`int_0^oo "e"^(- x/2) x^5  "d"x`


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


Choose the correct alternative:

`int_(-1)^1 x^3 "e"^(x^4)  "d"x` is


Choose the correct alternative:

Using the factorial representation of the gamma function, which of the following is the solution for the gamma function Γ(n) when n = 8 is


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


`int x^9/(4x^2 + 1)^6  "d"x` is equal to ______.


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


Evaluate \[\int_{1}^{3}(2x+1)\,dx\].


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×