English

Π / 2 ∫ 0 Cos 3 X D X - Mathematics

Advertisements
Advertisements

Question

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

Solution

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

shaalaa.com
Definite Integrals
  Is there an error in this question or solution?
Chapter 20: Definite Integrals - Exercise 20.1 [Page 16]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 20 Definite Integrals
Exercise 20.1 | Q 18 | Page 16

RELATED QUESTIONS

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

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

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

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

\[\int\limits_{- 1}^1 \frac{1}{x^2 + 2x + 5} dx\]

\[\int_0^\frac{\pi}{4} \left( \tan x + \cot x \right)^{- 2} dx\]

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

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

\[\int\limits_0^{\pi/2} \sqrt{\sin \phi} \cos^5 \phi\ d\phi\]

 


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

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

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

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

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

\[\int_{- \frac{\pi}{2}}^\pi \sin^{- 1} \left( \sin x \right)dx\]

If  \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that \[\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]

 


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

Evaluate the following integral:

\[\int_{- a}^a \log\left( \frac{a - \sin\theta}{a + \sin\theta} \right)d\theta\]

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

 


If f is an integrable function, show that

\[\int\limits_{- a}^a x f\left( x^2 \right) dx = 0\]

 


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

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

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

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

Evaluate each of the following integral:

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

 


Evaluate each of the following  integral:

\[\int_0^1 x e^{x^2} dx\]

 


The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .


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


\[\int\limits_{- 1}^1 \left| 1 - x \right| dx\]  is equal to

Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .

 

`int_0^(2a)f(x)dx`


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


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


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


\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]


\[\int\limits_2^3 e^{- x} dx\]


Evaluate the following:

`int_0^oo "e"^(- x/2) x^5  "d"x`


Evaluate the following integrals as the limit of the sum:

`int_1^3 x  "d"x`


`int (cos2x - cos 2theta)/(cosx - costheta) "d"x` is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×