मराठी

Π / 2 ∫ 0 X + Sin X 1 + Cos X D X

Advertisements
Advertisements

प्रश्न

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

उत्तर

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

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Definite Integrals - Exercise 20.2 [पृष्ठ ३९]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
Exercise 20.2 | Q 29 | पृष्ठ ३९

संबंधित प्रश्‍न

\[\int\limits_2^3 \frac{x}{x^2 + 1} dx\]

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

\[\int\limits_1^3 \frac{\log x}{\left( x + 1 \right)^2} dx\]

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

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

\[\int_0^\frac{\pi}{4} \left( \tan x + \cot x \right)^{- 2} dx\]

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

\[\int\limits_0^9 f\left( x \right) dx, where f\left( x \right) \begin{cases}\sin x & , & 0 \leq x \leq \pi/2 \\ 1 & , & \pi/2 \leq x \leq 3 \\ e^{x - 3} & , & 3 \leq x \leq 9\end{cases}\]

Evaluate the following integral:

\[\int\limits_{- 3}^3 \left| x + 1 \right| dx\]

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

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

\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx, 0 < \alpha < \pi\]

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

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

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

\[\int\limits_0^3 \left( 2 x^2 + 3x + 5 \right) dx\]

Solve each of the following integral:

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

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

\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^3 x} dx\]  is equal to

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

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

`int_0^(2a)f(x)dx`


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


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


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


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


\[\int\limits_0^{2\pi} \cos^7 x dx\]


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


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


\[\int\limits_2^3 e^{- x} dx\]


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


Evaluate the following:

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


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


Which notation represents a definite integral?


Which integral has Constant C present?


Evaluate \[\int_{0}^{1}x\,dx\].


What is an antiderivative of \[2x+1\]?


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×