Advertisements
Advertisements
प्रश्न
\[\int x \cos x\ dx\]
बेरीज
उत्तर
\[\int x \text{ cos x dx }\]
\[\text{Taking x as the first function and cos x as the second function} . \]
\[ = x\int\cos x dx - \int\left\{ \frac{d}{dx}\left( x \right)\int\text{ cos x dx }\right\}dx\]
\[ = x \sin x - \int\text{ sin x dx }\]
\[ = x \sin x + \cos x + C\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int \left( \tan x + \cot x \right)^2 dx\]
\[\int\frac{1 - \cos 2x}{1 + \cos 2x} dx\]
\[\int\frac{1}{1 - \sin x} dx\]
\[\int\frac{\left( x^3 + 8 \right)\left( x - 1 \right)}{x^2 - 2x + 4} dx\]
\[\int\frac{1}{2 - 3x} + \frac{1}{\sqrt{3x - 2}} dx\]
\[\int\frac{\cos 4x - \cos 2x}{\sin 4x - \sin 2x} dx\]
\[\int\frac{\left( x + 1 \right) e^x}{\cos^2 \left( x e^x \right)} dx\]
\[\int\frac{x}{\sqrt{x^2 + a^2} + \sqrt{x^2 - a^2}} dx\]
\[\int\frac{x^2 + 3x + 1}{\left( x + 1 \right)^2} dx\]
` ∫ 1 /{x^{1/3} ( x^{1/3} -1)} ` dx
\[\int\frac{dx}{e^x + e^{- x}}\]
\[\int\frac{1}{\sqrt{\left( x - \alpha \right)\left( \beta - x \right)}} dx, \left( \beta > \alpha \right)\]
\[\int\frac{\sin 2x}{\sqrt{\cos^4 x - \sin^2 x + 2}} dx\]
\[\int\frac{x + 1}{\sqrt{x^2 + 1}} dx\]
\[\int\frac{2}{2 + \sin 2x}\text{ dx }\]
\[\int\frac{2 \sin x + 3 \cos x}{3 \sin x + 4 \cos x} dx\]
\[\int \sin^{- 1} \left( 3x - 4 x^3 \right) \text{ dx }\]
\[\int x^2 \sin^{- 1} x\ dx\]
\[\int\frac{\left( x \tan^{- 1} x \right)}{\left( 1 + x^2 \right)^{3/2}} \text{ dx }\]
\[\int x^3 \tan^{- 1}\text{ x dx }\]
\[\int \cos^3 \sqrt{x}\ dx\]
\[\int \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}} dx\]
\[\int e^x \left( \frac{1 + \sin x}{1 + \cos x} \right) dx\]
\[\int\frac{e^x \left( x - 4 \right)}{\left( x - 2 \right)^3} \text{ dx }\]
\[\int\sqrt{3 - 2x - 2 x^2} \text{ dx}\]
\[\int\sqrt{x^2 - 2x} \text{ dx}\]
\[\int\left( x + 1 \right) \sqrt{2 x^2 + 3} \text{ dx}\]
\[\int\left( x + 2 \right) \sqrt{x^2 + x + 1} \text{ dx }\]
\[\int\left( x + 1 \right) \sqrt{x^2 + x + 1} \text{ dx }\]
\[\int\frac{x^2 + 1}{\left( 2x + 1 \right) \left( x^2 - 1 \right)} dx\]
\[\int\frac{x^2 + 1}{x^4 + x^2 + 1} \text{ dx }\]
\[\int\sqrt{\cot \text{θ} d } \text{ θ}\]
\[\int\frac{1}{\left( 1 + x^2 \right) \sqrt{1 - x^2}} \text{ dx }\]
\[\int\frac{x}{\left( x^2 + 4 \right) \sqrt{x^2 + 9}} \text{ dx}\]
\[\int\frac{\sin^6 x}{\cos^8 x} dx =\]
\[\int\frac{x^3}{x + 1}dx\] is equal to
\[\int\frac{\sin x + \cos x}{\sqrt{\sin 2x}} \text{ dx}\]
\[\int\frac{x^2}{\left( x - 1 \right)^3} dx\]
\[\int\frac{1}{\sin^2 x + \sin 2x} \text{ dx }\]
\[\int\frac{1}{\left( x^2 + 2 \right) \left( x^2 + 5 \right)} \text{ dx}\]
