हिंदी

∫ √ 2 a X − X 2 D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\sqrt{2ax - x^2} \text{ dx}\]
योग
Advertisements

उत्तर

\[\text{ Let I }= \int\sqrt{2ax - x^2}\text{ dx}\]
\[ = \int\sqrt{a^2 + 2ax - x^2 - a^2}\text{ dx}\]
\[ = \int \sqrt{a^2 - \left( x^2 - 2ax + a^2 \right)}\text{ dx}\]
\[ = \int\sqrt{a^2 - \left( x - a \right)^2}\text{ dx}\]
\[ = \left( \frac{x - a}{2} \right) \sqrt{2ax - x^2} + \frac{a^2}{2} \sin^{- 1} \left( \frac{x - a}{a} \right) + C\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Indefinite Integrals - Exercise 19.28 [पृष्ठ १५५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.28 | Q 15 | पृष्ठ १५५

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

\[\int \tan^2 \left( 2x - 3 \right) dx\]


\[\int\frac{x^2 + x + 5}{3x + 2} dx\]

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

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

` ∫  tan 2x tan 3x  tan 5x    dx  `

\[\int\left\{ 1 + \tan x \tan \left( x + \theta \right) \right\} dx\]

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

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

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

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

\[\int\frac{x}{3 x^4 - 18 x^2 + 11} dx\]

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

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

\[\int\frac{x}{\sqrt{8 + x - x^2}} dx\]


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

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

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

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

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

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

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

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

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

Evaluate the following integral:

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

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

\[\int\frac{1}{\cos x + \sqrt{3} \sin x} \text{ dx } \] is equal to

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

\[\int\frac{\left( 2^x + 3^x \right)^2}{6^x} \text{ dx }\] 

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

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

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

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

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

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

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

Evaluate : \[\int\frac{\cos 2x + 2 \sin^2 x}{\cos^2 x}dx\] .


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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×