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

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

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

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

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

\[\int\limits_1^3 \frac{\cos \left( \log x \right)}{x} dx\]

\[\int\limits_0^{\pi/2} \frac{dx}{a \cos x + b \sin x}a, b > 0\]

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

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

\[\int_{- \frac{\pi}{2}}^\pi \sin^{- 1} \left( \sin x \right)dx\]

Evaluate each of the following integral:

\[\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx, n \in N, n \geq 2\]


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

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

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

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

 


If f is an integrable function, show that

\[\int\limits_{- a}^a x f\left( x^2 \right) dx = 0\]

 


Prove that:

\[\int_0^\pi xf\left( \sin x \right)dx = \frac{\pi}{2} \int_0^\pi f\left( \sin x \right)dx\]

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

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

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

\[\int\limits_{- \pi/2}^{\pi/2} x \cos^2 x\ dx .\]

 


Evaluate each of the following  integral:

\[\int_0^1 x e^{x^2} dx\]

 


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


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


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

 


\[\int\limits_0^1 \frac{d}{dx}\left\{ \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \right\} dx\] is equal to

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


\[\int\limits_0^\pi x \sin x \cos^4 x dx\]


\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]


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


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


Using second fundamental theorem, evaluate the following:

`int_0^(1/4) sqrt(1 - 4)  "d"x`


Evaluate the following:

`int_0^2 "f"(x)  "d"x` where f(x) = `{{:(3 - 2x - x^2",", x ≤ 1),(x^2 + 2x - 3",", 1 < x ≤ 2):}`


Evaluate the following:

`int_0^oo "e"^(- x/2) x^5  "d"x`


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 + 4)  "d"x`


Choose the correct alternative:

`int_0^1 (2x + 1)  "d"x` is


Choose the correct alternative:

Γ(1) is


`int x^3/(x + 1)` is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×