मराठी

∫ X Sin 2 X D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int x \text{ sin 2x dx }\]
बेरीज
Advertisements

उत्तर

\[\int x \text{ sin 2x dx }\]
` "Taking x as the first function and sin 2x as the second function  " `.
\[ = x\int\text{ sin 2x dx} - \int\left\{ \frac{d}{dx}\left( x \right)\int\text{ sin 2x dx }\right\}dx\]
\[ = \frac{- x \cos 2x}{2} + \int\frac{\cos 2x}{2}dx\]
\[ = \frac{- x \cos 2x}{2} + \frac{\sin 2x}{4} + C\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Indefinite Integrals - Exercise 19.25 [पृष्ठ १३३]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Exercise 19.25 | Q 9 | पृष्ठ १३३

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

\[\int\sqrt{x}\left( x^3 - \frac{2}{x} \right) dx\]

If f' (x) = x − \[\frac{1}{x^2}\]  and  f (1)  \[\frac{1}{2},    find  f(x)\]

 


If f' (x) = x + bf(1) = 5, f(2) = 13, find f(x)


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

\[\int \cos^2 \text{nx dx}\]

\[\int x^3 \sin x^4 dx\]

\[\int x^2 e^{x^3} \cos \left( e^{x^3} \right) dx\]

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

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

\[\int\frac{e^{m \tan^{- 1} x}}{1 + x^2} dx\]

` ∫    \sqrt{tan x}     sec^4  x   dx `


\[\int\frac{1}{\sqrt{5 x^2 - 2x}} dx\]

` ∫  {x-3} /{ x^2 + 2x - 4 } dx `


\[\int\frac{2x + 5}{x^2 - x - 2} \text{ dx }\]

\[\int\frac{x + 7}{3 x^2 + 25x + 28}\text{ dx}\]

\[\int\frac{x^2}{x^2 + 6x + 12} \text{ dx }\]

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

\[\int\frac{1}{4 \sin^2 x + 5 \cos^2 x} \text{ dx }\]

\[\int x e^x \text{ dx }\]

\[\int\frac{x + \sin x}{1 + \cos x} \text{ dx }\]

\[\int \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]

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

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

\[\int\frac{\sqrt{16 + \left( \log x \right)^2}}{x} \text{ dx}\]

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

\[\int\frac{18}{\left( x + 2 \right) \left( x^2 + 4 \right)} dx\]

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

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

\[\int\frac{x}{\left( x - 3 \right) \sqrt{x + 1}} dx\]

\[\int\left( x - 1 \right) e^{- x} dx\] is equal to

\[\int\frac{1}{1 + \tan x} dx =\]

\[\int\frac{1}{\left( \sin^{- 1} x \right) \sqrt{1 - x^2}} \text{ dx} \]

\[\int\text{ cos x  cos  2x   cos  3x  dx}\]


\[\int\frac{\sin x}{\sqrt{1 + \sin x}} dx\]

\[\int\frac{x^3}{\sqrt{x^8 + 4}} \text{ dx }\]


\[\int\sqrt{\frac{a + x}{x}}dx\]
 

\[\int \tan^5 x\ \sec^3 x\ dx\]

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


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

\[\int\frac{x^2}{x^2 + 7x + 10} dx\]

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×