English

Π / 2 ∫ 0 Sin X √ 1 + Cos X D X - Mathematics

Advertisements
Advertisements

Question

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

Sum

Solution

\[\int_0^\frac{\pi}{2} \frac{\sin x}{\sqrt{1 + \cos x}} d x\]

\[Let 1 + \cos x = t, then - \sin x dx = dt\]

\[When, x \to 0, t \to 2 and x \to \frac{\pi}{2}, t \to 1\]

Therefore, the integral becomes

\[ \int_2^1 \frac{- 1}{\sqrt{t}}dt\]

\[ = \int_1^2 \frac{1}{\sqrt{t}}dt\]

\[ = 2 \left[ \sqrt{t} \right]_1^2 \]

\[ = 2\left( \sqrt{2} - 1 \right)\]

shaalaa.com
Definite Integrals
  Is there an error in this question or solution?
Chapter 20: Definite Integrals - Revision Exercise [Page 121]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 20 Definite Integrals
Revision Exercise | Q 12 | Page 121

RELATED QUESTIONS

\[\int\limits_{\pi/3}^{\pi/4} \left( \tan x + \cot x \right)^2 dx\]

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

 


\[\int\limits_0^1 \frac{2x + 3}{5 x^2 + 1} dx\]

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

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

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

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

\[\int\limits_1^2 \frac{3x}{9 x^2 - 1} dx\]

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

\[\int\limits_0^1 \frac{24 x^3}{\left( 1 + x^2 \right)^4} dx\]

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

\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]


\[\int\limits_0^2 \left( x^2 + 1 \right) dx\]

\[\int\limits_2^3 \left( 2 x^2 + 1 \right) dx\]

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

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

\[\int\limits_0^2 \left( x^2 + 2x + 1 \right) dx\]

\[\int\limits_2^3 x^2 dx\]

Solve each of the following integral:

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

If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]


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

\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals


\[\int\limits_0^3 \frac{3x + 1}{x^2 + 9} dx =\]

If \[I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx,\]  then the value of I10 + 90I8 is

 


Evaluate : \[\int e^{2x} \cdot \sin \left( 3x + 1 \right) dx\] .


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


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


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


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


\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]


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


\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]


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_(- pi/4)^(pi/4) x^3 cos^3 x  "d"x`


Evaluate the following:

`Γ (9/2)`


Evaluate the following integrals as the limit of the sum:

`int_0^1 (x + 4)  "d"x`


Find `int sqrt(10 - 4x + 4x^2)  "d"x`


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×