English

Π / 4 ∫ 0 Cos 4 X Sin 3 X D X - Mathematics

Advertisements
Advertisements

Question

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

Sum

Solution

\[\int_0^\frac{\pi}{4} \cos^4 x \sin^3 x d x\]

\[ = \int_0^\frac{\pi}{4} \cos^4 x \sin x \left( 1 - \cos^2 x \right) dx\]

\[ = \int_0^\frac{\pi}{4} \cos^4 x \sin x dx - \int_0^\frac{\pi}{4} \cos^6 x \sin x dx\]

\[ = - \left[ \frac{\cos^5 x}{5} \right]_0^\frac{\pi}{4} + \left[ \frac{\cos^7 x}{7} \right]_0^\frac{\pi}{4} \]

\[ = \frac{- 1}{20\sqrt{2}} + \frac{1}{5} + \frac{1}{56\sqrt{2}} - \frac{1}{7}\]

\[ = \frac{- \sqrt{2}}{40} + \frac{2}{35} + \frac{\sqrt{2}}{112}\]

\[ = \frac{2}{35} - \frac{9\sqrt{2}}{560}\]

shaalaa.com
Definite Integrals
  Is there an error in this question or solution?
Chapter 20: Definite Integrals - Revision Exercise [Page 121]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 20 Definite Integrals
Revision Exercise | Q 19 | Page 121

RELATED QUESTIONS

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

\[\int\limits_0^{\pi/6} \cos x \cos 2x\ dx\]

Evaluate the following definite integrals:

\[\int_0^\frac{\pi}{2} x^2 \sin\ x\ dx\]

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

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

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

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

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

\[\int\limits_2^4 \frac{x}{x^2 + 1} dx\]

\[\int\limits_0^{\pi/2} \frac{\sin \theta}{\sqrt{1 + \cos \theta}} d\theta\]

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

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

 


Evaluate each of the following integral:

\[\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx, n \in N, n \geq 2\]


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

If `f` is an integrable function such that f(2a − x) = f(x), then prove that

\[\int\limits_0^{2a} f\left( x \right) dx = 2 \int\limits_0^a f\left( x \right) dx\]

 


If f(2a − x) = −f(x), prove that

\[\int\limits_0^{2a} f\left( x \right) dx = 0 .\]

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

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

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

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

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

Evaluate each of the following  integral:

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

 


Evaluate each of the following integral:

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

 


If \[\left[ \cdot \right] and \left\{ \cdot \right\}\] denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:

\[\int\limits_0^{\pi/4} \sin \left\{ x \right\} dx\]

 


Given that \[\int\limits_0^\infty \frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)\left( x^2 + c^2 \right)} dx = \frac{\pi}{2\left( a + b \right)\left( b + c \right)\left( c + a \right)},\] the value of \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]


\[\int\limits_0^3 \frac{3x + 1}{x^2 + 9} dx =\]

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


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


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


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


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


\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]


\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]


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


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


Evaluate the following using properties of definite integral:

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


Choose the correct alternative:

`int_(-1)^1 x^3 "e"^(x^4)  "d"x` is


Choose the correct alternative:

If f(x) is a continuous function and a < c < b, then `int_"a"^"c" f(x)  "d"x + int_"c"^"b" f(x)  "d"x` is


Choose the correct alternative:

If n > 0, then Γ(n) is


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×