English

Π / 2 ∫ 0 Cos 4 X D X

Advertisements
Advertisements

Question

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

 

Sum
Advertisements

Solution

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

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Definite Integrals - Exercise 20.1 [Page 16]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 19 Definite Integrals
Exercise 20.1 | Q 22 | Page 16

RELATED QUESTIONS

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

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

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

\[\int\limits_0^{\pi/2} \frac{dx}{a \cos x + b \sin x}a, b > 0\]

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

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

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

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

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

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

Evaluate the following integral:

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

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

\[\int\limits_0^1 \left( 3 x^2 + 5x \right) dx\]

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

\[\int\limits_0^2 \left( x^2 + 2x + 1 \right) dx\]

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

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

\[\int\limits_0^1 e^\left\{ x \right\} dx .\]

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


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

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

The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is

 


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

 


The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is 

 


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


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


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


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


Using second fundamental theorem, evaluate the following:

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


Evaluate the following using properties of definite integral:

`int_0^1 x/((1 - x)^(3/4))  "d"x`


Choose the correct alternative:

`int_0^1 (2x + 1)  "d"x` is


Choose the correct alternative:

Γ(n) is


Choose the correct alternative:

Γ(1) is


Choose the correct alternative:

`int_0^oo x^4"e"^-x  "d"x` is


Evaluate `int sqrt((1 + x)/(1 - x)) "d"x`, x ≠1


Evaluate the following:

`int ((x^2 + 2))/(x + 1) "d"x`


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


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


What is an antiderivative of \[x\]?


What are definite integrals used to find over a fixed interval?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×