हिंदी

∫ X Sin X Cos X D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int x \sin x \cos x\ dx\]

 

योग
Advertisements

उत्तर

\[\int x\sin x \cdot \text{ cos x dx }\]
\[ = \frac{1}{2}\int x\left( 2 \sin x \cos x \right) dx\]
\[ = \frac{1}{2}\int x_{} \cdot \sin \left( 2x \right)_{} dx\]
\[\text{Taking x as the first function and sin 2x as the second function} . \]
\[ = \frac{1}{2}\left[ x\int\text{ sin 2x dx } - \int\left\{ \frac{d}{dx}\left( x \right)\int\text{ sin 2x dx } \right\}dx \right]\]
\[ = \frac{1}{2}\left[ x \times \frac{- \text{ cos }\left( 2x \right)}{2} - \int1 \cdot \left( \frac{- \cos 2x}{2} \right)dx \right]\]
\[ = \frac{1}{2}\left[ \frac{- x \text{ cos
}\left( 2x \right)}{2} + \frac{\text{ sin } \left( 2x \right)}{4} \right] + C\]
\[ = \frac{- x \text{ cos } \left( 2x \right)}{4} + \frac{\text{ sin }\left( 2x \right)}{8} + C\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Indefinite Integrals - Exercise 19.25 [पृष्ठ १३३]

APPEARS IN

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

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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

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

\[\int\left( 5x + 3 \right) \sqrt{2x - 1} dx\]

` ∫  {sin 2x} /{a cos^2  x  + b sin^2  x }  ` dx 


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

\[\int\frac{\cos x - \sin x}{1 + \sin 2x} dx\]

\[\int\frac{\sin 2x}{\left( a + b \cos 2x \right)^2} dx\]

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

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

` ∫  tan^3    x   sec^2  x   dx  `

` ∫   tan   x   sec^4  x   dx  `


` ∫  tan^5 x   sec ^4 x   dx `

\[\int \cot^5 \text{ x } {cosec}^4 x\text{ dx }\]

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

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

\[\int\frac{a x^3 + bx}{x^4 + c^2} dx\]

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

\[\int\frac{5x + 3}{\sqrt{x^2 + 4x + 10}} \text{ dx }\]

\[\int\frac{1}{1 - \cot x} dx\]

\[\int\frac{1}{4 + 3 \tan x} dx\]

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

\[\int e^x \left( \frac{\sin 4x - 4}{1 - \cos 4x} \right) dx\]

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

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

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

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

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

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

\[\int\frac{x^2 + 1}{x^4 + 7 x^2 + 1} 2 \text{ dx }\]

\[\int\frac{\left( x - 1 \right)^2}{x^4 + x^2 + 1} \text{ dx}\]

` \int \text{ x} \text{ sec x}^2 \text{  dx  is  equal  to }`

 


If \[\int\frac{\sin^8 x - \cos^8 x}{1 - 2 \sin^2 x \cos^2 x} dx\]


\[\int e^x \left( 1 - \cot x + \cot^2 x \right) dx =\]

\[\int\frac{1}{1 - \cos x - \sin x} dx =\]

\[\int\frac{x^9}{\left( 4 x^2 + 1 \right)^6}dx\]  is equal to 

\[\int\frac{x^3}{x + 1}dx\] is equal to

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

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

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


\[\int\sqrt{1 + 2x - 3 x^2}\text{  dx } \]

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×