मराठी

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

Advertisements
Advertisements

प्रश्न

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

बेरीज
Advertisements

उत्तर

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

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

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

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

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

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

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

shaalaa.com
Definite Integrals
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 20: Definite Integrals - Revision Exercise [पृष्ठ १२१]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 20 Definite Integrals
Revision Exercise | Q 19 | पृष्ठ १२१

संबंधित प्रश्‍न

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

\[\int\limits_0^2 \frac{1}{\sqrt{3 + 2x - x^2}} dx\]

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

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

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

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

\[\int_\frac{1}{3}^1 \frac{\left( x - x^3 \right)^\frac{1}{3}}{x^4}dx\]

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

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

Evaluate the following integral:

\[\int\limits_{- 3}^3 \left| x + 1 \right| dx\]

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

 


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

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

If f (x) is a continuous function defined on [0, 2a]. Then, prove that

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

 


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

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

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

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

 


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

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

Evaluate each of the following integral:

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

 


If \[\int_0^a \frac{1}{4 + x^2}dx = \frac{\pi}{8}\] , find the value of a.


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


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


\[\int\limits_0^{\pi/2} \sin\ 2x\ \log\ \tan x\ dx\]  is equal to 

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


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


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


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


\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]


Using second fundamental theorem, evaluate the following:

`int_0^(1/4) sqrt(1 - 4)  "d"x`


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):}`


Evaluate the following:

`Γ (9/2)`


Evaluate the following integrals as the limit of the sum:

`int_1^3 x  "d"x`


Choose the correct alternative:

`int_0^oo "e"^(-2x)  "d"x` is


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:

`int_0^oo x^4"e"^-x  "d"x` is


If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.


Verify the following:

`int (2x + 3)/(x^2 + 3x) "d"x = log|x^2 + 3x| + "C"`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×