हिंदी

∫ X 2 Cos 2 X D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int x^2 \cos 2x\ \text{ dx }\]
योग

उत्तर

\[\int x^2 \text{ cos 2x dx }\]
`  " Taking x"^2 " as the first function and cos 2x as the second function" .` 
\[ = x^2 \int\text{ cos 2x dx } - \int\left( 2x\int\text{ cos 2x dx }\right)dx\]
\[ = \frac{x^2 \sin 2x}{2} - \int\frac{2x \sin 2x}{2}dx\]
\[ = \frac{x^2}{2}\sin 2x - \int x \text{ sin 2x dx }\]
\[ = \frac{x^2}{2}\sin 2x - \left[ x\int\sin2x - \int\left( \int\text{ sin 2x dx }\right)dx \right]\]
\[ = \frac{x^2}{2}\sin 2x - \left[ \frac{- x \cos 2x}{2} + \int\frac{\cos 2x}{2}dx \right]\]
\[ = \frac{x^2}{2}\sin 2x + \frac{x \cos 2x}{2} - \frac{\sin 2x}{4} + C\]

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 8 | पृष्ठ १३३

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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

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

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

` ∫  1/ {1+ cos   3x}  ` dx


\[\int     \text{sin}^2  \left( 2x + 5 \right)    \text{dx}\]

`∫     cos ^4  2x   dx `


\[\int\text{sin mx }\text{cos nx dx m }\neq n\]

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

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

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

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

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

\[\int\frac{\sec^2 x}{1 - \tan^2 x} dx\]

\[\int\frac{e^x}{e^{2x} + 5 e^x + 6} dx\]

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

\[\int\frac{1}{\sqrt{16 - 6x - x^2}} dx\]

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

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

\[\int\frac{1}{5 + 7 \cos x + \sin x} dx\]

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

\[\int \sec^{- 1} \sqrt{x}\ dx\]

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

\[\int e^x \left[ \sec x + \log \left( \sec x + \tan x \right) \right] dx\]

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

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

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

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

\[\int(2x + 5)\sqrt{10 - 4x - 3 x^2}dx\]

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

\[\int\frac{x^2 + 6x - 8}{x^3 - 4x} dx\]

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

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

If \[\int\frac{1}{5 + 4 \sin x} dx = A \tan^{- 1} \left( B \tan\frac{x}{2} + \frac{4}{3} \right) + C,\] then


If \[\int\frac{2^{1/x}}{x^2} dx = k 2^{1/x} + C,\]  then k is equal to


\[\int\frac{\sin^2 x}{\cos^4 x} dx =\]

\[\int \sin^4 2x\ dx\]

\[\int\frac{6x + 5}{\sqrt{6 + x - 2 x^2}} \text{ dx}\]

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

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


Find: `int (3x +5)/(x^2+3x-18)dx.`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×