English

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

Advertisements
Advertisements

Question

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

 

Sum
Advertisements

Solution

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

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

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

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

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

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

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

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

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

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

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

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

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Definite Integrals - Exercise 20.5 [Page 94]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 19 Definite Integrals
Exercise 20.5 | Q 5 | Page 94

RELATED QUESTIONS

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

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

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

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


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

\[\int\limits_1^2 \log\ x\ dx\]

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

\[\int_0^1 \frac{1}{1 + 2x + 2 x^2 + 2 x^3 + x^4}dx\]

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

\[\int\limits_0^{\pi/2} \frac{\sin \theta}{\sqrt{1 + \cos \theta}} d\theta\]

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

\[\int\limits_0^{\pi/4} \sin^3 2t \cos 2t\ dt\]

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

\[\int\limits_0^{\pi/6} \cos^{- 3} 2 \theta \sin 2\ \theta\ d\ \theta\]

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

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

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

\[\int\limits_0^7 \frac{\sqrt[3]{x}}{\sqrt[3]{x} + \sqrt[3]{7} - x} dx\]

If f(x) is a continuous function defined on [−aa], then prove that 

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

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

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

If \[\int\limits_0^1 \left( 3 x^2 + 2x + k \right) dx = 0,\] find the value of k.

 


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

\[\int\limits_{- 1}^1 \left| 1 - x \right| dx\]  is equal to

The value of \[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\] is

 


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

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


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


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


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


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


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


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


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


Using second fundamental theorem, evaluate the following:

`int_(-1)^1 (2x + 3)/(x^2 + 3x + 7)  "d"x`


Choose the correct alternative:

`int_(-1)^1 x^3 "e"^(x^4)  "d"x` is


Choose the correct alternative:

Γ(1) is


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


If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×