Advertisements
Advertisements
Question
Advertisements
Solution
\[\int2x \sec^3 \left( x^2 + 3 \right) \cdot \tan \left( x^2 + 3 \right) dx\]
\[ = \int \sec^2 \left( x^2 + 3 \right) \cdot \sec \left( x^2 + 3 \right) \cdot \tan \left( x^2 + 3 \right) \cdot \text{2x dx}\]
\[\text{Let }\sec \left( x^2 + 3 \right) = t\]
\[ \Rightarrow \sec \left( x^2 + 3 \right) \cdot \tan \left( x^2 + 3 \right) \cdot 2x = \frac{dt}{dx}\]
\[ \Rightarrow \sec \left( x^2 + 3 \right) \cdot \tan \left( x^2 + 3 \right) \cdot \text{2x dx} = dt\]
\[Now, \int \sec^2 \left( x^2 + 3 \right) \cdot \sec \left( x^2 + 3 \right) \cdot \tan \left( x^2 + 3 \right) \cdot \text{2x dx}\]
\[ = \int t^2 dt\]
\[ = \frac{t^3}{3} + C\]
\[ = \frac{\sec^3 \left( x^2 + 3 \right)}{3} + C\]
APPEARS IN
RELATED QUESTIONS
\[\int\frac{x}{\sqrt{8 + x - x^2}} dx\]
\[\int {cosec}^4 2x\ dx\]
\[ \int\left( 1 + x^2 \right) \ \cos 2x \ dx\]
