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प्रश्न
\[\int2x \sec^3 \left( x^2 + 3 \right) \tan \left( x^2 + 3 \right) dx\]
योग
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उत्तर
\[\int2x \sec^3 \left( x^2 + 3 \right) \cdot \tan \left( x^2 + 3 \right) dx\]
\[ = \int \sec^2 \left( x^2 + 3 \right) \cdot \sec \left( x^2 + 3 \right) \cdot \tan \left( x^2 + 3 \right) \cdot \text{2x dx}\]
\[\text{Let }\sec \left( x^2 + 3 \right) = t\]
\[ \Rightarrow \sec \left( x^2 + 3 \right) \cdot \tan \left( x^2 + 3 \right) \cdot 2x = \frac{dt}{dx}\]
\[ \Rightarrow \sec \left( x^2 + 3 \right) \cdot \tan \left( x^2 + 3 \right) \cdot \text{2x dx} = dt\]
\[Now, \int \sec^2 \left( x^2 + 3 \right) \cdot \sec \left( x^2 + 3 \right) \cdot \tan \left( x^2 + 3 \right) \cdot \text{2x dx}\]
\[ = \int t^2 dt\]
\[ = \frac{t^3}{3} + C\]
\[ = \frac{\sec^3 \left( x^2 + 3 \right)}{3} + C\]
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