मराठी

Π / 2 ∫ 0 Sin N X Sin N X + Cos N X D X

Advertisements
Advertisements

प्रश्न

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

 

बेरीज
Advertisements

उत्तर

\[Let\ I = \int_0^\frac{\pi}{2} \frac{\sin^n x}{\sin^n x + \cos^n x} d x ..............(1)\]

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

\[ = \int_0^\frac{\pi}{2} \frac{\cos^n x}{\cos^n x + \sin^n x} dx \]

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

\[\text{Adding (1) and (2) we get}\]

\[2I = \int_0^\frac{\pi}{2} \frac{\sin^n x}{\sin^n x + \cos^n x} + \frac{\cos^n x}{\sin^n x + \cos^n x} d x \]

\[ = \int_0^\frac{\pi}{2} \frac{\sin^n x + \cos^n x}{\sin^n x + \cos^n x} dx\]

\[ = \int_0^\frac{\pi}{2} 1\ dx\]

\[ = \int_0^\frac{\pi}{2} dx\]

\[ = \left[ x \right]_0^\frac{\pi}{2} \]

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

\[Hence\ I = \frac{\pi}{4}\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Definite Integrals - Exercise 20.5 [पृष्ठ ९४]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
Exercise 20.5 | Q 5 | पृष्ठ ९४

संबंधित प्रश्‍न

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

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

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

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

\[\int\limits_0^{\pi/4} \frac{\tan^3 x}{1 + \cos 2x} dx\]

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

\[\int_0^\frac{1}{2} \frac{1}{\left( 1 + x^2 \right)\sqrt{1 - x^2}}dx\]

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

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

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

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

\[\int\limits_0^2 x\sqrt{2 - x} dx\]

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

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

 


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

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

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

\[\int\limits_0^{15} \left[ x \right] dx .\]

`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`


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

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


If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to


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


Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .


\[\int\limits_0^4 x\sqrt{4 - x} dx\]


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


Evaluate the following integrals :-

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


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


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


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


\[\int\limits_0^4 x dx\]


Evaluate the following:

`int_(-1)^1 "f"(x)  "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x  < 0):}`


Evaluate the following:

`Γ (9/2)`


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


Evaluate the following integrals as the limit of the sum:

`int_0^1 x^2  "d"x`


Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`


Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`


What is the meaning of a definite integral?


Evaluate \[\int_{1}^{3}(2x+1)\,dx\].


What are definite integrals used to find over a fixed interval?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×