हिंदी

∫ X Cos 2 X D X - Mathematics

Advertisements
Advertisements

प्रश्न

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

उत्तर

\[\int x \cos^2 x dx\]
`  " Taking x as the first function and cos"^2 x " as the second function ." `

\[ = x\int\frac{1 + \cos 2x}{2}dx - \int\left\{ \frac{d}{dx}\left( x \right)\int\frac{1 + \cos 2x}{2}dx \right\}dx\]
\[ = \frac{x}{2}\left[ x + \frac{\sin2x}{2} \right] - \int\frac{1}{2}\left( x + \frac{\sin2x}{2} \right)dx\]
\[ = \frac{x}{2}\left[ x + \frac{\sin2x}{2} \right] - \left[ \frac{x^2}{4} - \frac{\cos2x}{8} \right] + C\]
\[ = \frac{x^2}{2} + \frac{x \sin2x}{2} - \frac{x^2}{4} + \frac{\cos2x}{8} + C\]
\[ = \frac{x^2}{4} + \frac{x \sin2x}{2} + \frac{\cos2x}{8} + C\]

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 13 | पृष्ठ १३३

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

\[\int\frac{\left( x^3 + 8 \right)\left( x - 1 \right)}{x^2 - 2x + 4} dx\]

If f' (x) = a sin x + b cos x and f' (0) = 4, f(0) = 3, f

\[\left( \frac{\pi}{2} \right)\] = 5, find f(x)
 

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

\[\int 5^{5^{5^x}} 5^{5^x} 5^x dx\]

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

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

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

\[\int\frac{1}{4 x^2 + 12x + 5} dx\]

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

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

\[\int\frac{1}{\sqrt{5 - 4x - 2 x^2}} dx\]

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

\[\int\frac{x - 1}{3 x^2 - 4x + 3} dx\]

\[\int\frac{1 - 3x}{3 x^2 + 4x + 2}\text{  dx}\]

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

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

\[\int\frac{3x + 1}{\sqrt{5 - 2x - x^2}} \text{ dx }\]

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

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

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

`int"x"^"n"."log"  "x"  "dx"`

\[\int\frac{\log x}{x^n}\text{  dx }\]

\[\int {cosec}^3 x\ dx\]

\[\int \cos^{- 1} \left( 4 x^3 - 3x \right) \text{ dx }\]

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

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

\[\int e^x \left( \cot x - {cosec}^2 x \right) dx\]

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


\[\int\frac{\sin^6 x}{\cos^8 x} dx =\]

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

\[\int \sin^5 x\ dx\]

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

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

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

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

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

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

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

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

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×