Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\int x^3 \cos x^2 \text{ dx }\]
\[\text{ Let x}^2 = t \]
\[ \Rightarrow 2x = \frac{dt}{dx}\]
\[ \Rightarrow dx = \frac{dt}{2x}\]
\[ = \frac{1}{2}\left[ \int t \text{ cos t dt } \right]\]
\[\text{Taking t as the first function and cos t as the second function} . \]
\[ = \frac{1}{2}\left[ t\sin t - \int\text{ sin t dt } \right]\]
\[ = \frac{1}{2}\left[ t\sin t + \cos t \right] . . . (1) \]
\[\text{Substituting the value of t in eq} \text{ (1) } \]
\[ = \frac{x^2 \sin x^2}{2} + \frac{\cos x^2}{2} + c\]
APPEARS IN
संबंधित प्रश्न
If f' (x) = 8x3 − 2x, f(2) = 8, find f(x)
` ∫ sin x \sqrt (1-cos 2x) dx `
Write a value of
\[\int\frac{1}{\sqrt{x} + \sqrt{x + 1}} \text{ dx }\]
\[\int\frac{1 - x^4}{1 - x} \text{ dx }\]
\[\int\frac{x^3}{\sqrt{x^8 + 4}} \text{ dx }\]
\[\int \left( e^x + 1 \right)^2 e^x dx\]
