Advertisements
Advertisements
प्रश्न
\[\int\frac{x^2}{\sqrt{1 - x}} \text{ dx }\]
बेरीज
Advertisements
उत्तर
\[\text{We have}, \]
\[I = \int\frac{x^2}{\sqrt{1 - x}} dx\]
\[\text{ Let, 1 - x = t}^2 \]
\[\text{Differentiating both sides we get}\]
\[ - \text{ dx = 2t dt}\]
\[\text{Now, integration becomes}, \]
\[I = - \int\frac{\left( 1 - t^2 \right)^2}{t} 2tdt\]
\[ = - 2\int \left( 1 - t^2 \right)^2 dt\]
\[ = - 2\int\left( 1 - 2 t^2 + t^4 \right) dt\]
\[ = - 2\left[ t - \frac{2 t^3}{3} + \frac{t^5}{5} \right] + C\]
\[ = \frac{- 2}{15}t\left[ 3 t^4 - 10 t^2 + 15 \right] + C\]
\[ = \frac{- 2}{15}\sqrt{1 - x}\left[ 3 \left( 1 - x \right)^2 - 10\left( 1 - x \right) + 15 \right] + C\]
\[ = \frac{- 2}{15}\sqrt{1 - x}\left[ 3\left( 1 - 2x + x^2 \right) - 10\left( 1 - x \right) + 15 \right] + C\]
\[ = \frac{- 2}{15}\sqrt{1 - x}\left[ 3 x^2 - 6x + 3 - 10 + 10x + 15 \right] + C\]
\[ = \frac{- 2}{15}\sqrt{1 - x}\left[ 3 x^2 + 4x + 8 \right] + C\]
\[I = \int\frac{x^2}{\sqrt{1 - x}} dx\]
\[\text{ Let, 1 - x = t}^2 \]
\[\text{Differentiating both sides we get}\]
\[ - \text{ dx = 2t dt}\]
\[\text{Now, integration becomes}, \]
\[I = - \int\frac{\left( 1 - t^2 \right)^2}{t} 2tdt\]
\[ = - 2\int \left( 1 - t^2 \right)^2 dt\]
\[ = - 2\int\left( 1 - 2 t^2 + t^4 \right) dt\]
\[ = - 2\left[ t - \frac{2 t^3}{3} + \frac{t^5}{5} \right] + C\]
\[ = \frac{- 2}{15}t\left[ 3 t^4 - 10 t^2 + 15 \right] + C\]
\[ = \frac{- 2}{15}\sqrt{1 - x}\left[ 3 \left( 1 - x \right)^2 - 10\left( 1 - x \right) + 15 \right] + C\]
\[ = \frac{- 2}{15}\sqrt{1 - x}\left[ 3\left( 1 - 2x + x^2 \right) - 10\left( 1 - x \right) + 15 \right] + C\]
\[ = \frac{- 2}{15}\sqrt{1 - x}\left[ 3 x^2 - 6x + 3 - 10 + 10x + 15 \right] + C\]
\[ = \frac{- 2}{15}\sqrt{1 - x}\left[ 3 x^2 + 4x + 8 \right] + C\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{\sin^3 x - \cos^3 x}{\sin^2 x \cos^2 x} dx\]
\[\int \sin^{- 1} \left( \frac{2 \tan x}{1 + \tan^2 x} \right) dx\]
\[\int\frac{1}{\sqrt{x + a} + \sqrt{x + b}} dx\]
\[\int\frac{1 + \cos x}{1 - \cos x} dx\]
\[\int \tan^2 \left( 2x - 3 \right) dx\]
\[\int\frac{x + 1}{\sqrt{2x + 3}} dx\]
\[\int \text{sin}^2 \left( 2x + 5 \right) \text{dx}\]
` ∫ sin 4x cos 7x dx `
\[\int\frac{1}{\sqrt{1 + \cos x}} dx\]
\[\int\frac{e^{3x}}{e^{3x} + 1} dx\]
\[\int\frac{1 - \cot x}{1 + \cot x} dx\]
\[\int\frac{\left( 1 + \sqrt{x} \right)^2}{\sqrt{x}} dx\]
\[\int {cosec}^4 \text{ 3x } \text{ dx } \]
\[\int \cos^5 x \text{ dx }\]
\[\int\frac{1}{\sqrt{8 + 3x - x^2}} dx\]
\[\int\frac{x - 1}{3 x^2 - 4x + 3} dx\]
\[\int\frac{6x - 5}{\sqrt{3 x^2 - 5x + 1}} \text{ dx }\]
\[\int x^3 \text{ log x dx }\]
\[\int\left( x + 1 \right) \text{ e}^x \text{ log } \left( x e^x \right) dx\]
\[\int\frac{\left( x \tan^{- 1} x \right)}{\left( 1 + x^2 \right)^{3/2}} \text{ dx }\]
\[\int \tan^{- 1} \left( \sqrt{x} \right) \text{dx }\]
\[\int\frac{x^3 \sin^{- 1} x^2}{\sqrt{1 - x^4}} \text{ dx }\]
\[\int e^x \left( \tan x - \log \cos x \right) dx\]
\[\int\frac{x^2 \sin^{- 1} x}{\left( 1 - x^2 \right)^{3/2}} \text{ dx }\]
\[\int e^x \left( \frac{\sin 4x - 4}{1 - \cos 4x} \right) dx\]
\[\int\left\{ \tan \left( \log x \right) + \sec^2 \left( \log x \right) \right\} dx\]
\[\int x\sqrt{x^4 + 1} \text{ dx}\]
\[\int\frac{x}{\left( x + 1 \right) \left( x^2 + 1 \right)} dx\]
\[\int\frac{x^4}{\left( x - 1 \right) \left( x^2 + 1 \right)} dx\]
Evaluate the following integral:
\[\int\frac{x^2}{1 - x^4}dx\]
\[\int\frac{\left( x - 1 \right)^2}{x^4 + x^2 + 1} \text{ dx}\]
\[\int\frac{1}{\left( x + 1 \right) \sqrt{x^2 + x + 1}} \text{ dx }\]
\[\int\frac{x^4 + x^2 - 1}{x^2 + 1} \text{ dx}\]
\[\int\sin x \sin 2x \text{ sin 3x dx }\]
\[\int\frac{1}{\text{ sin} \left( x - a \right) \text{ sin } \left( x - b \right)} \text{ dx }\]
\[\int \tan^5 x\ dx\]
\[\int\frac{1}{2 + \cos x} \text{ dx }\]
\[\int x\sqrt{1 + x - x^2}\text{ dx }\]
\[\int x^3 \left( \log x \right)^2\text{ dx }\]
