Advertisements
Advertisements
प्रश्न
\[\int\frac{1}{\sqrt{1 - x^2}\left( 2 + 3 \sin^{- 1} x \right)} dx\]
बेरीज
Advertisements
उत्तर
\[\text{Let I} = \int\frac{1}{\sqrt{1 - x^2}\left( 2 + 3 \sin^{- 1} x \right)}dx\]
\[\text{Putting}\ \sin^{- 1} x = t\]
\[ \Rightarrow \frac{1}{\sqrt{1 - x^2}} = \frac{dt}{dx}\]
\[ \Rightarrow \frac{1}{\sqrt{1 - x^2}}dx = dt\]
\[ \therefore I = \int\frac{1}{2 + 3t}dt\]
\[ = \frac{1}{3} \text{ln }\left| 2 + 3t \right| + C\]
\[ = \frac{1}{3} \text{ln }\left| 2 + 3 \sin^{- 1} x \right| + C \left[ \because t = \sin^{- 1} x \right]\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{\left( 1 + \sqrt{x} \right)^2}{\sqrt{x}} dx\]
\[\int\frac{1 - \cos x}{1 + \cos x} dx\]
\[\int\frac{x^2 + 5x + 2}{x + 2} dx\]
`∫ cos ^4 2x dx `
\[\int\sqrt{\frac{1 + \cos 2x}{1 - \cos 2x}} dx\]
\[\int\sqrt{\frac{1 - \sin 2x}{1 + \sin 2x}} dx\]
\[\int\frac{\sec^2 x}{\tan x + 2} dx\]
\[\int\frac{\cos x - \sin x}{1 + \sin 2x} dx\]
\[\int\frac{\left( x + 1 \right) e^x}{\cos^2 \left( x e^x \right)} dx\]
\[\int2x \sec^3 \left( x^2 + 3 \right) \tan \left( x^2 + 3 \right) dx\]
\[\int 5^{x + \tan^{- 1} x} . \left( \frac{x^2 + 2}{x^2 + 1} \right) dx\]
\[\int\left( 2 x^2 + 3 \right) \sqrt{x + 2} \text{ dx }\]
` ∫ tan^3 x sec^2 x dx `
\[\int \cot^5 x \text{ dx }\]
\[\int \sin^5 x \cos x \text{ dx }\]
\[\int\frac{1}{\sqrt{\left( 2 - x \right)^2 + 1}} dx\]
\[\int\frac{x}{x^4 + 2 x^2 + 3} dx\]
\[\int\frac{x}{\sqrt{x^4 + a^4}} dx\]
\[\int\frac{\cos x - \sin x}{\sqrt{8 - \sin2x}}dx\]
\[\int\frac{1 - 3x}{3 x^2 + 4x + 2}\text{ dx}\]
\[\int x^2 \sin^2 x\ dx\]
\[\int2 x^3 e^{x^2} dx\]
\[\int \sin^{- 1} \sqrt{x} \text{ dx }\]
\[\int x^3 \tan^{- 1}\text{ x dx }\]
\[\int\frac{\sqrt{1 - \sin x}}{1 + \cos x} e^{- x/2} \text{ dx }\]
\[\int e^x \cdot \frac{\sqrt{1 - x^2} \sin^{- 1} x + 1}{\sqrt{1 - x^2}} \text{ dx }\]
∴\[\int e^{2x} \left( - \sin x + 2 \cos x \right) dx\]
\[\int\frac{1}{\left( x - 1 \right) \left( x + 1 \right) \left( x + 2 \right)} dx\]
\[\int\frac{x^2 + x + 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} dx\]
\[\int\frac{1}{\left( x^2 + 1 \right) \sqrt{x}} \text{ dx }\]
\[\int\frac{1}{\left( 2 x^2 + 3 \right) \sqrt{x^2 - 4}} \text{ dx }\]
If \[\int\frac{\sin^8 x - \cos^8 x}{1 - 2 \sin^2 x \cos^2 x} dx\]
\[\int\frac{\cos 2x - 1}{\cos 2x + 1} dx =\]
\[\int\frac{x + 2}{\left( x + 1 \right)^3} \text{ dx }\]
\[\int\sqrt{\sin x} \cos^3 x\ \text{ dx }\]
\[\int\frac{1}{\sqrt{x^2 - a^2}} \text{ dx }\]
\[\int\frac{1}{4 x^2 + 4x + 5} dx\]
\[\int\frac{1}{2 + \cos x} \text{ dx }\]
\[\int x^3 \left( \log x \right)^2\text{ dx }\]
\[\int\sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} \text{ dx}\]
