Question

Evaluate the following integrals. $$\text{Missing left or extra right}$$

Short answer

Question: Evaluate the double integral $${\iint }_{R}x{\sec }^{2}ydA$$, where R is the region defined by $(x,y):0\le y\le x^2,0\le x\le \sqrt{\pi }/2$. Answer: The double integral is equal to $$\frac{1}{2}\ln (\sqrt{2})$$.

Step-by-step solution

Identifying the limits of integration

Given the region R, we can see that the limits of integration for y are from 0 to $x^2$, and for x, the limits are from 0 to $\sqrt{\pi }/2$.

Evaluate the inner integral

We start by integrating the given function with respect to y: $${\int }_0^{x^2}x{\sec }^{2}ydy$$ Since x is a constant with respect to y, we can take it out of the integral: $$x{\int }_0^{x^2}{\sec }^{2}ydy$$ Now, we integrate it with respect to y: $$x{[\tan y]}_0^{x^2}=x(\tan (x^2)-\tan (0))$$ The inner integral simplifies to: $$x\tan (x^2)$$

Evaluate the outer integral

Now, we have to integrate our result with respect to x: $${\int }_0^{\sqrt{\pi }/2}x\tan (x^2)dx$$ To evaluate this integral, we will use substitution. Let $u=x^2$. Then, $\frac{du}{dx}=2x$, or $dx=\frac{du}{2x}$. The bounds of integration will also change: When $x=0$, $u=0$. When $x=\sqrt{\pi }/2$, $u=\pi /4$. Our integral now becomes: $${\int }_0^{\pi /4}\tan (u)\frac{du}{2}=\frac{1}{2}{\int }_0^{\pi /4}\tan (u)du$$

Integrating with respect to u

Now we just need to integrate $\tan (u)$ with respect to u: $$\frac{1}{2}{[-\ln (\cos (u))]}_0^{\pi /4}=\frac{1}{2}(-\ln (\cos (\pi /4))+\ln (\cos (0)))$$ We know that $\cos (\pi /4)=\frac{\sqrt{2}}{2}$ and $\cos (0)=1$. So, our result is: $$\frac{1}{2}(-\ln (\frac{\sqrt{2}}{2})+\ln (1))=\frac{1}{2}(\ln (2)-\ln (\sqrt{2}))$$

Simplifying the result

By using the properties of logarithms, we can simplify the result: $$\frac{1}{2}(\ln (2)-\ln (\sqrt{2}))=\frac{1}{2}(\ln \frac{2}{\sqrt{2}})=\frac{1}{2}\ln \left(\frac{2\sqrt{2}}{2}\right)$$ $$=\frac{1}{2}\ln (\sqrt{2})$$ So the integral evaluates to: $${\iint }_{R}x{\sec }^{2}ydA=\frac{1}{2}\ln (\sqrt{2})$$