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प्रश्न
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उत्तर
\[\int x^3 \cos x^2 \text{ dx }\]
\[\text{ Let x}^2 = t \]
\[ \Rightarrow 2x = \frac{dt}{dx}\]
\[ \Rightarrow dx = \frac{dt}{2x}\]
\[ = \frac{1}{2}\left[ \int t \text{ cos t dt } \right]\]
\[\text{Taking t as the first function and cos t as the second function} . \]
\[ = \frac{1}{2}\left[ t\sin t - \int\text{ sin t dt } \right]\]
\[ = \frac{1}{2}\left[ t\sin t + \cos t \right] . . . (1) \]
\[\text{Substituting the value of t in eq} \text{ (1) } \]
\[ = \frac{x^2 \sin x^2}{2} + \frac{\cos x^2}{2} + c\]
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संबंधित प्रश्न
` = ∫ root (3){ cos^2 x} sin x dx `
\[\int x\ {cosec}^2 \text{ x }\ \text{ dx }\]
The value of \[\int\frac{\sin x + \cos x}{\sqrt{1 - \sin 2x}} dx\] is equal to
\[\int\frac{x^3}{\sqrt{1 + x^2}}dx = a \left( 1 + x^2 \right)^\frac{3}{2} + b\sqrt{1 + x^2} + C\], then
\[\int\frac{x + 2}{\left( x + 1 \right)^3} \text{ dx }\]
