हिंदी

Π ∫ 0 5 ( 5 − 4 Cos θ ) 1 / 4 Sin θ D θ

Advertisements
Advertisements

प्रश्न

\[\int\limits_0^\pi 5 \left( 5 - 4 \cos \theta \right)^{1/4} \sin \theta\ d \theta\]
Advertisements

उत्तर

\[Let\ I = \int_0^\pi 5 \left( 5 - 4\cos \theta \right)^\frac{1}{4} \sin \theta\ d \theta . \]

\[Let\left( 5 - 4 \cos \theta \right) = t . Then, 4 \sin \theta\ d\theta = dt\]

\[When\ \theta = 0, t = 1\ and\ \theta = \pi, t = 9\]

\[ \therefore I = \frac{5}{4} \int_1^9 t^\frac{1}{4} dt\]

\[ \Rightarrow I = \frac{5}{4} \left[ \frac{4 t^\frac{5}{4}}{5} \right]_1^9 \]

\[ \Rightarrow I = \left( 9\sqrt{3} - 1 \right)\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Definite Integrals - Exercise 20.2 [पृष्ठ ३९]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 19 Definite Integrals
Exercise 20.2 | Q 42 | पृष्ठ ३९

संबंधित प्रश्न

\[\int\limits_{- 2}^3 \frac{1}{x + 7} dx\]

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

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

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

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

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

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

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

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

\[\int_0^\frac{1}{2} \frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}dx\]

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

\[\int\limits_0^1 \frac{1 - x^2}{x^4 + x^2 + 1} dx\]

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

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

\[\int\limits_0^\pi x \cos^2 x\ dx\]

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

Evaluate the following integral:

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

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

\[\int\limits_0^2 e^x dx\]

\[\int\limits_1^2 \log_e \left[ x \right] dx .\]

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

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

 


If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals

 


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

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


\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]


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


\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]


\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]


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


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


Evaluate the following using properties of definite integral:

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


If f(x) = `{{:(x^2"e"^(-2x)",", x ≥ 0),(0",", "otherwise"):}`, then evaluate `int_0^oo "f"(x) "d"x`


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


Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:


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


What is the result of a definite integral?


What is the meaning of a definite integral?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×