English

Π / 2 ∫ 0 Cos 5 X D X

Advertisements
Advertisements

Question

\[\int\limits_0^{\pi/2} \cos^5 x\ dx\]
Advertisements

Solution

\[Let\ I = \int_0^\frac{\pi}{2} \cos^5 x\ d\ x . Then, \]
\[I = \int_0^\frac{\pi}{2} \cos^4 x \cos\ x\ d x\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} \left( 1 - \sin^2 x \right)^2 \cos x\ dx\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} \left( 1 - 2 \sin^2 x + \sin^4 x \right) \cos x dx \]
\[Let\ \sin x = t . Then, \cos dx = du\]
\[When\ x = 0, t = 0\ and\ x = \frac{\pi}{2}, t = 1\]
\[ \therefore I = \int_0^1 \left( 1 - 2 t^2 + t^4 \right) dt\]
\[ \Rightarrow I = \left[ t - \frac{2 t^3}{3} + \frac{t^5}{5} \right]_0^1 \]
\[ \Rightarrow I = 1 - \frac{2}{3} + \frac{1}{5}\]
\[ \Rightarrow I = \frac{8}{15}\]
shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Definite Integrals - Exercise 20.2 [Page 40]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 19 Definite Integrals
Exercise 20.2 | Q 46 | Page 40

RELATED QUESTIONS

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

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

\[\int\limits_1^3 \frac{\log x}{\left( x + 1 \right)^2} dx\]

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

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

\[\int_0^\frac{\pi}{4} \left( a^2 \cos^2 x + b^2 \sin^2 x \right)dx\]

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

\[\int_0^\frac{1}{2} \frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}dx\]

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

\[\int_0^\frac{\pi}{2} \frac{\cos^2 x}{1 + 3 \sin^2 x}dx\]

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

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

\[\int_0^\frac{\pi}{4} \frac{\sin^2 x \cos^2 x}{\left( \sin^3 x + \cos^3 x \right)^2}dx\]

\[\int\limits_0^9 f\left( x \right) dx, where f\left( x \right) \begin{cases}\sin x & , & 0 \leq x \leq \pi/2 \\ 1 & , & \pi/2 \leq x \leq 3 \\ e^{x - 3} & , & 3 \leq x \leq 9\end{cases}\]

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

 


\[\int_0^2 2x\left[ x \right]dx\]

Evaluate each of the following integral:

\[\int_0^{2\pi} \log\left( \sec x + \tan x \right)dx\]

 


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

 


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

 


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

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

Evaluate the following integral:

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

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

\[\int\limits_0^2 e^x dx\]

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

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

Evaluate each of the following integral:

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

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

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


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


\[\int\limits_0^1 \left| 2x - 1 \right| dx\]


\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]


\[\int\limits_0^1 \left| \sin 2\pi x \right| dx\]


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


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


Evaluate the following using properties of definite integral:

`int_(- pi/2)^(pi/2) sin^2theta  "d"theta`


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


Choose the correct alternative:

If n > 0, then Γ(n) is


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×