मराठी

∫ X 2 √ a 6 − X 6 D X - Mathematics

Advertisements
Advertisements

प्रश्न

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

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
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Indefinite Integrals - Exercise 19.28 [पृष्ठ १५४]

APPEARS IN

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

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

` ∫  {cosec x} / {"cosec x "- cot x} ` dx      


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

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


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


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

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

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

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

\[\int \sec^4 2x \text{ dx }\]

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

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

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

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

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

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

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

`  ∫ \sqrt{"cosec x"- 1}  dx `

\[\int\frac{\left( 3 \sin x - 2 \right) \cos x}{5 - \cos^2 x - 4 \sin x} dx\]

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

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

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

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

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

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

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

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

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

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

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

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

\[\int\sqrt{\cot \text{θ} d  } \text{ θ}\]

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

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

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

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

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

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


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

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

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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×