English

Π ∫ 0 X Sin X Cos 4 X D X

Advertisements
Advertisements

Question

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

Sum
Advertisements

Solution

\[Let, I = \int_0^\pi x \sin x \cos^4 x d x ............(1)\]

\[ = \int_0^\pi \left( \pi - x \right) \sin\left( \pi - x \right) \cos^4 \left( \pi - x \right) d x \]

\[ = \int_0^\pi \left( \pi - x \right) \sin x \cos^4 x d x ..............(2)\]

Adding (1) and (2)

\[2I = \int_0^\pi \left[ x \sin x \cos^4 x + \left( \pi - x \right) \sin x \cos^4 x \right] d x \]

\[ = \int_0^\pi \left( x + \pi - x \right) \sin x \cos^4 x d x \]

\[ = \pi \int_0^\pi \sin x \cos^4 x d x \]

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

\[ = \pi\left[ \frac{1}{5} + \frac{1}{5} \right]\]

\[ = \frac{2\pi}{5}\]

\[Hence, I = \frac{\pi}{5}\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Definite Integrals - Revision Exercise [Page 122]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 19 Definite Integrals
Revision Exercise | Q 42 | Page 122

RELATED QUESTIONS

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

\[\int\limits_{\pi/6}^{\pi/4} cosec\ x\ dx\]

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

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

\[\int\limits_1^e \frac{\log x}{x} dx\]

\[\int\limits_1^4 \frac{x^2 + x}{\sqrt{2x + 1}} dx\]

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

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

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

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

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

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

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

\[\int_{- 2}^2 x e^\left| x \right| dx\]

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

Evaluate each of the following integral:

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


\[\int\limits_{- \pi/4}^{\pi/4} \sin^2 x\ dx\]

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

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

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

\[\int\limits_0^4 \frac{1}{\sqrt{16 - x^2}} dx .\]

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

The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is 


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 \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]


The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]

 


\[\lim_{n \to \infty} \left\{ \frac{1}{2n + 1} + \frac{1}{2n + 2} + . . . + \frac{1}{2n + n} \right\}\] is equal to

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


\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]


\[\int\limits_1^3 \left| x^2 - 2x \right| dx\]


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


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


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


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


Evaluate the following:

`int_0^2 "f"(x)  "d"x` where f(x) = `{{:(3 - 2x - x^2",", x ≤ 1),(x^2 + 2x - 3",", 1 < x ≤ 2):}`


Choose the correct alternative:

`int_0^1 (2x + 1)  "d"x` is


Choose the correct alternative:

`Γ(3/2)`


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


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


`int (x + 3)/(x + 4)^2 "e"^x  "d"x` = ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×