Advertisements
Advertisements
प्रश्न
\[\int x^2 \text{ cos x dx }\]
योग
उत्तर
\[\int x^2 \text{ cos x dx }\]
` "Taking x"^2" as the first function and cos x as the second function " . `
\[ = x^2 \int\cos x dx - \int\left( \frac{d}{dx} x^2 \int\text{ cos x dx } \right)dx\]
\[ = x^2 \sin x - \int2x \text{ sin x dx }\]
\[ = x^2 \sin x - 2\left[ x\int\sin x - \int\left\{ \frac{d}{dx}\left( x \right)\int\text{ sin x dx } \right\}dx \right]\]
\[ = x^2 \sin x - 2\left[ - x\cos x + \int\text{ cos x dx } \right]\]
\[ = x^2 \sin x + 2x \cos x - 2 \sin x + C\]
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\left( 2 - 3x \right) \left( 3 + 2x \right) \left( 1 - 2x \right) dx\]
\[\int\left( \frac{m}{x} + \frac{x}{m} + m^x + x^m + mx \right) dx\]
\[\int\frac{\sin^2 x}{1 + \cos x} \text{dx} \]
\[\int \left( a \tan x + b \cot x \right)^2 dx\]
\[\int\frac{\sin 2x}{\left( a + b \cos 2x \right)^2} dx\]
\[\int2x \sec^3 \left( x^2 + 3 \right) \tan \left( x^2 + 3 \right) dx\]
\[\int\frac{\sin \left( \text{log x} \right)}{x} dx\]
` ∫ tan^5 x sec ^4 x dx `
\[\int \cos^5 x \text{ dx }\]
\[\int\frac{1}{\sin x \cos^3 x} dx\]
\[\int\frac{\cos x}{\sqrt{4 + \sin^2 x}} dx\]
\[\int\frac{1}{4 \sin^2 x + 5 \cos^2 x} \text{ dx }\]
\[\int\frac{\sin 2x}{\sin^4 x + \cos^4 x} \text{ dx }\]
\[\int\frac{1}{3 + 2 \sin x + \cos x} \text{ dx }\]
\[\int\frac{1}{1 - \tan x} \text{ dx }\]
`int"x"^"n"."log" "x" "dx"`
\[\int\frac{\log x}{x^n}\text{ dx }\]
\[\int x \sin x \cos x\ dx\]
\[\int \cos^3 \sqrt{x}\ dx\]
\[\int\left\{ \tan \left( \log x \right) + \sec^2 \left( \log x \right) \right\} dx\]
\[\int\frac{2x}{\left( x^2 + 1 \right) \left( x^2 + 3 \right)} dx\]
\[\int\frac{x^2 + x - 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} dx\]
\[\int\frac{x}{\left( x + 1 \right) \left( x^2 + 1 \right)} dx\]
Evaluate the following integral:
\[\int\frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)}dx\]
\[\int\frac{x^4}{\left( x - 1 \right) \left( x^2 + 1 \right)} dx\]
\[\int\frac{\sin x}{\cos 2x} \text{ dx }\]
\[\int \tan^4 x\ dx\]
\[\int \sin^3 x \cos^4 x\ \text{ dx }\]
\[\int\frac{1}{x^2 + 4x - 5} \text{ dx }\]
\[\int\frac{\sin x}{\sqrt{\cos^2 x - 2 \cos x - 3}} \text{ dx }\]
\[\int\frac{1}{\sin^4 x + \cos^4 x} \text{ dx}\]
\[\int {cosec}^4 2x\ dx\]
\[\int\sqrt{\frac{a + x}{x}}dx\]
\[\int\sqrt{a^2 + x^2} \text{ dx }\]
\[\int \left( x + 1 \right)^2 e^x \text{ dx }\]
\[\int x\sqrt{\frac{1 - x}{1 + x}} \text{ dx }\]
\[\int\frac{x^2 + x + 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} \text{ dx}\]
\[\int \sin^3 \left( 2x + 1 \right) \text{dx}\]
Find: `int (sin2x)/sqrt(9 - cos^4x) dx`
