हिंदी

∫ X 2 √ a 6 − X 6 D X - Mathematics

Advertisements
Advertisements

प्रश्न

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

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

\[\int\frac{x^6 + 1}{x^2 + 1} dx\]

\[\int \left( a \tan x + b \cot x \right)^2 dx\]

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

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

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

\[\int\frac{1 + \cos 4x}{\cot x - \tan x} dx\]

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

\[\int\frac{3x + 5}{\sqrt{7x + 9}} dx\]

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

\[\int\frac{x}{\sqrt{x + a} - \sqrt{x + b}}dx\]

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

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

\[\int x^3 \sin x^4 dx\]

\[\int\frac{\cos\sqrt{x}}{\sqrt{x}} dx\]

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

\[\int x \cos^3 x^2 \sin x^2 \text{ dx }\]

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

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

` ∫  {x-3} /{ x^2 + 2x - 4 } dx `


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

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

\[\int x^2 \text{ cos x dx }\]

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

\[\int e^x \left( \tan x - \log \cos x \right) dx\]

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

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

\[\int(2x + 5)\sqrt{10 - 4x - 3 x^2}dx\]

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

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

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


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

\[\int\frac{\cos2x - \cos2\theta}{\cos x - \cos\theta}dx\] is equal to 

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

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

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

\[\int\left( 2x + 3 \right) \sqrt{4 x^2 + 5x + 6} \text{ dx}\]

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

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

\[\int\frac{5 x^4 + 12 x^3 + 7 x^2}{x^2 + x} dx\]


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×