मराठी

∫ √ 3 − X 2 D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\sqrt{3 - x^2} \text{ dx}\]
बेरीज
Advertisements

उत्तर

\[\text{ Let I } = \int\sqrt{3 - x^2}\text{ dx}\]
\[ = \int\sqrt{\left( \sqrt{3} \right)^2 - x^2}\text{ dx} \]
\[ = \frac{x}{2}\sqrt{\left( \sqrt{3} \right)^2 - x^2} + \frac{\left( \sqrt{3} \right)^2}{2} \sin^{- 1} \left( \frac{x}{\sqrt{3}} \right) + C\]
\[ = \frac{x}{2} \sqrt{3 - x^2} + \frac{3}{2} \sin^{- 1} \left( \frac{x}{\sqrt{3}} \right) + C\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Indefinite Integrals - Exercise 19.28 [पृष्ठ १५५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Exercise 19.28 | Q 16 | पृष्ठ १५५

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

\[\int\left( 3x\sqrt{x} + 4\sqrt{x} + 5 \right)dx\]

\[\int\frac{2 x^4 + 7 x^3 + 6 x^2}{x^2 + 2x} dx\]

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

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

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


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

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

\[\int \tan^{3/2} x \sec^2 \text{x dx}\]

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

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

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

\[\int \sin^4 x \cos^3 x \text{ dx }\]

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

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

\[\int\frac{e^{3x}}{4 e^{6x} - 9} dx\]

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

\[\int\frac{1}{\sqrt{8 + 3x - x^2}} dx\]

\[\int\frac{1}{\sqrt{\left( 1 - x^2 \right)\left\{ 9 + \left( \sin^{- 1} x \right)^2 \right\}}} dx\]

`  ∫ \sqrt{"cosec x"- 1}  dx `

\[\int\frac{\cos x - \sin x}{\sqrt{8 - \sin2x}}dx\]

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

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

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

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

`int"x"^"n"."log"  "x"  "dx"`

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

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

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

\[\int\frac{e^x \left( x - 4 \right)}{\left( x - 2 \right)^3} \text{ dx }\]

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

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

\[\int\frac{x}{\left( x^2 + 4 \right) \sqrt{x^2 + 9}} \text{ dx}\]

Write the anti-derivative of  \[\left( 3\sqrt{x} + \frac{1}{\sqrt{x}} \right) .\]


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

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

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

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

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

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

\[\int\frac{1}{\left( x^2 + 2 \right) \left( x^2 + 5 \right)} \text{ dx}\]

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×