Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\int x^2 \text{ cos 2x dx }\]
` " Taking x"^2 " as the first function and cos 2x as the second function" .`
\[ = x^2 \int\text{ cos 2x dx } - \int\left( 2x\int\text{ cos 2x dx }\right)dx\]
\[ = \frac{x^2 \sin 2x}{2} - \int\frac{2x \sin 2x}{2}dx\]
\[ = \frac{x^2}{2}\sin 2x - \int x \text{ sin 2x dx }\]
\[ = \frac{x^2}{2}\sin 2x - \left[ x\int\sin2x - \int\left( \int\text{ sin 2x dx }\right)dx \right]\]
\[ = \frac{x^2}{2}\sin 2x - \left[ \frac{- x \cos 2x}{2} + \int\frac{\cos 2x}{2}dx \right]\]
\[ = \frac{x^2}{2}\sin 2x + \frac{x \cos 2x}{2} - \frac{\sin 2x}{4} + C\]
APPEARS IN
संबंधित प्रश्न
If f' (x) = x − \[\frac{1}{x^2}\] and f (1) \[\frac{1}{2}, find f(x)\]
` ∫ sin x \sqrt (1-cos 2x) dx `
Write a value of
If `int(2x^(1/2))/(x^2) dx = k . 2^(1/x) + C`, then k is equal to ______.
\[\int\frac{1 - x^4}{1 - x} \text{ dx }\]
\[\int {cosec}^4 2x\ dx\]
