English

Π / 2 ∫ 0 X + Sin X 1 + Cos X D X

Advertisements
Advertisements

Question

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

Solution

\[Let\, I = \int\limits_0^{\pi/2} \frac{x + \sin x}{1 + \cos x} dx\]
\[ = \int_0^{\pi/2} \frac{x + \sin x}{2 \cos^2 \frac{x}{2}} dx\]
\[ = \int_0^{\pi/2} \left[ \frac{x}{2 \cos^2 \frac{x}{2}} + \frac{\sin x}{2 \cos^2 \frac{x}{2}} \right]dx\]
\[ = \frac{1}{2} \int_0^{\pi/2} x se c^2 \frac{x}{2}dx + \int_0^{\pi/2} \frac{2\sin \frac{x}{2}\cos\frac{x}{2}}{2 \cos^2 \frac{x}{2}}dx\]
\[ = \frac{1}{2} \left[ x \frac{\tan\frac{x}{2}}{\frac{1}{2}} \right]_0^{\pi/2} - \frac{1}{2} \int_0^{\pi/2} \frac{\tan\frac{x}{2}}{\frac{1}{2}}dx + \int_0^{\pi/2} \tan\frac{x}{2}dx\]
\[ = \left[ x \tan\frac{x}{2} \right]_0^{\pi/2} - \int_0^{\pi/2} \tan\frac{x}{2} dx + \int_0^{\pi/2} \tan\frac{x}{2}dx\]
\[ = \left[ \frac{\pi}{2} \tan\frac{\pi}{4} \right]\]
\[ = \frac{\pi}{2} \times 1\]
\[ = \frac{\pi}{2} \]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Definite Integrals - Exercise 20.2 [Page 39]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 19 Definite Integrals
Exercise 20.2 | Q 29 | Page 39

RELATED QUESTIONS

\[\int\limits_4^9 \frac{1}{\sqrt{x}} dx\]

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

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

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

Evaluate the following definite integrals:

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

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

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

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

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

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

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

\[\int_0^\frac{\pi}{4} \frac{\sin x + \cos x}{3 + \sin2x}dx\]

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

\[\int_0^\frac{\pi}{4} \frac{\sin^2 x \cos^2 x}{\left( \sin^3 x + \cos^3 x \right)^2}dx\]

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

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

 


\[\int\limits_0^\pi x \sin^3 x\ dx\]

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

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

Evaluate each of the following integral:

\[\int_0^\frac{\pi}{4} \tan\ xdx\]

 


Evaluate each of the following integral:

\[\int_e^{e^2} \frac{1}{x\log x}dx\]

If \[\left[ \cdot \right] and \left\{ \cdot \right\}\] denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:

\[\int\limits_0^{\pi/4} \sin \left\{ x \right\} dx\]

 


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


\[\int\limits_0^{\pi/2} \frac{\cos x}{\left( 2 + \sin x \right)\left( 1 + \sin x \right)} dx\] equals

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

 


\[\int\limits_{- \pi/2}^{\pi/2} \sin\left| x \right| dx\]  is equal to

The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is

 


\[\int\limits_0^1 \frac{x}{\left( 1 - x \right)^\frac{5}{4}} dx =\]

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


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


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


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


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


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


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


Evaluate the following:

`int_0^oo "e"^(-mx) x^6 "d"x`


Evaluate the following integrals as the limit of the sum:

`int_1^3 x  "d"x`


`int "e"^x ((1 - x)/(1 + x^2))^2  "d"x` is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×