हिंदी

Π / 4 ∫ 0 Cos 4 X Sin 3 X D X

Advertisements
Advertisements

प्रश्न

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

योग
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
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Definite Integrals - Revision Exercise [पृष्ठ १२१]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 19 Definite Integrals
Revision Exercise | Q 19 | पृष्ठ १२१

संबंधित प्रश्न

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

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

Evaluate the following definite integrals:

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

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

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

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

\[\int_\frac{\pi}{6}^\frac{\pi}{3} \left( \tan x + \cot x \right)^2 dx\]

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

\[\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{- \frac{\pi}{2}}{\sqrt{\cos x \sin^2 x}}dx\]

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

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

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

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

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

 


\[\int\limits_1^2 \left( x^2 - 1 \right) dx\]

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

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

Solve each of the following integral:

\[\int_2^4 \frac{x}{x^2 + 1}dx\]

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


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

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

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

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


\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]


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


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


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


\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]


\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]


Using second fundamental theorem, evaluate the following:

`int_0^3 ("e"^x "d"x)/(1 + "e"^x)`


Evaluate the following using properties of definite integral:

`int_0^1 log (1/x - 1)  "d"x`


Evaluate the following integrals as the limit of the sum:

`int_0^1 (x + 4)  "d"x`


Choose the correct alternative:

Γ(n) is


Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x


Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`


Find `int sqrt(10 - 4x + 4x^2)  "d"x`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×