हिंदी

∫ X 2 √ a 6 − X 6 D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int x^2 \sqrt{a^6 - x^6} \text{ dx}\]
योग

उत्तर

\[\text{ Let I } = \int x^2 \sqrt{a^6 - x^6}\text{ \text{ dx}}\]
\[ = \int x^2 \sqrt{\left( a^3 \right)^2 - \left( x^3 \right)^2}\text{ dx}\]
\[Putting\ x^3 = t\]
\[ \Rightarrow 3 x^2 dx = dt\]
\[ \Rightarrow x^2 dx = \frac{dt}{3}\]
\[ \therefore I = \frac{1}{3}\int\sqrt{\left( a^3 \right)^2 - t^2}dt\]
\[ = \frac{1}{3} \left[ \frac{t}{2}\sqrt{\left( a^3 \right)^2 - t^2} + \frac{\left( a^3 \right)^2}{2} \text{ sin}^{- 1} \left( \frac{t}{a^3} \right) \right] + C\]
\[ = \frac{x^3}{6} \sqrt{a^6 - x^6} + \frac{a^6}{6} \text{ sin}^{- 1} \left( \frac{x^3}{a^3} \right) + C\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Indefinite Integrals - Exercise 19.28 [पृष्ठ १५४]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.28 | Q 13 | पृष्ठ १५४

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

\[\int\left( 2 - 3x \right) \left( 3 + 2x \right) \left( 1 - 2x \right) dx\]

\[\int\frac{5 \cos^3 x + 6 \sin^3 x}{2 \sin^2 x \cos^2 x} dx\]

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

\[\int\frac{\cos x}{\cos \left( x - a \right)} dx\] 

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

\[\int\sqrt{1 + e^x} .  e^x dx\]

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

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

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

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

\[\int \cos^7 x \text{ dx  } \]

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

\[\int\frac{1}{x^2 - 10x + 34} dx\]

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

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

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

\[\int\frac{1}{x \left( x^6 + 1 \right)} dx\]

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

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

\[\int\frac{\cos x}{\sqrt{\sin^2 x - 2 \sin x - 3}} dx\]

` ∫  {x-3} /{ x^2 + 2x - 4 } dx `


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

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

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

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

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

\[\int e^x \left( \cot x + \log \sin x \right) dx\]

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

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

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

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

The primitive of the function \[f\left( x \right) = \left( 1 - \frac{1}{x^2} \right) a^{x + \frac{1}{x}} , a > 0\text{ is}\]


\[\int\sqrt{\frac{x}{1 - x}} dx\]  is equal to


\[\int\frac{1}{\text{ sin} \left( x - a \right) \text{ sin } \left( x - b \right)} \text{ dx }\]

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


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

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

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

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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×