Question

Integration of tan root x sec square root x/root x dx

Answer 1

tan^2(√x) + C

Answer 2

The integral to be evaluated is: $\int \frac{\tan \sqrt{x} \sec^2 \sqrt{x}}{\sqrt{x}} dx$

We use the method of substitution. Let $u = \tan \sqrt{x}$.

Now, we need to find $du$ by differentiating $u$ with respect to $x$. $\frac{du}{dx} = \frac{d}{dx} (\tan \sqrt{x})$ Using the chain rule, $\frac{d}{dx} (\tan f(x)) = \sec^2 f(x) \cdot f'(x)$, where $f(x) = \sqrt{x}$. First, find $f'(x)$: $\frac{d}{dx} (\sqrt{x}) = \frac{d}{dx} (x^{1/2}) = \frac{1}{2} x^{1/2 - 1} = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}}$ Now, substitute this back into the derivative of $u$: $\frac{du}{dx} = \sec^2 \sqrt{x} \cdot \frac{1}{2\sqrt{x}}$ Rearranging to find $du$: $du = \frac{\sec^2 \sqrt{x}}{2\sqrt{x}} dx$ This implies: $2 du = \frac{\sec^2 \sqrt{x}}{\sqrt{x}} dx$

Now, substitute $u$ and $2du$ into the original integral: The integral can be rewritten as: $\int \left( \tan \sqrt{x} \right) \left( \frac{\sec^2 \sqrt{x}}{\sqrt{x}} dx \right)$ Substitute $u = \tan \sqrt{x}$ and $2du = \frac{\sec^2 \sqrt{x}}{\sqrt{x}} dx$: $\int u (2 du)$ $= 2 \int u \, du$ Now, integrate with respect to $u$: $= 2 \left( \frac{u^{1+1}}{1+1} \right) + C$ $= 2 \left( \frac{u^2}{2} \right) + C$ $= u^2 + C$ Finally, substitute back $u = \tan \sqrt{x}$: $= (\tan \sqrt{x})^2 + C$ $= \tan^2 \sqrt{x} + C$

The final answer is $\tan^2 \sqrt{x} + C$.

Explanation of the solution: The integral is solved using the substitution method. By letting $u = \tan \sqrt{x}$, its differential $du = \sec^2 \sqrt{x} \cdot \frac{1}{2\sqrt{x}} dx$ perfectly matches the remaining part of the integrand (up to a constant factor). This transforms the integral into a simple power rule integration in terms of $u$, which is then easily evaluated and converted back to the original variable $x$.

Answer: The integral is $\tan^2 \sqrt{x} + C$.