MHB Evaluation of definite integral (is it correct?)

[Noah1]

To find the intersection(s) in the given interval, we may write:

$$\sin(x)=\cos(x)$$

Now, observing that $$\cos\left(\frac{\pi}{2}\right)=\cos\left(\frac{3\pi}{2}\right)=0$$ and:

$$\cos\left(\frac{\pi}{2}\right)\ne\sin\left(\frac{\pi}{2}\right)$$

$$\cos\left(\frac{3\pi}{2}\right)\ne\sin\left(\frac{3\pi}{2}\right)$$

We may then divide though by $\cos(x)$ without losing any solutions on the given interval to get:

$$\tan(x)=1$$

And, given the periodicity of the tangent function and the quadrant I solution of $$x=\frac{\pi}{4}$$, we may give the general solution as:

$$x=\frac{\pi}{4}+k\pi=\frac{\pi}{4}(4k+1)$$ where $k\in\mathbb{Z}$

Now, along with the given interval and what we found at the endpoints, we may write:

$$\frac{\pi}{2}<\frac{\pi}{4}(4k+1)<\frac{3\pi}{2}$$

$$2<4k+1<6$$

$$1<4k<5$$

$$\frac{1}{4}<k<\frac{5}{4}$$

Hence (given that $k$ is an integer):

$k=1$

And so the intersection of the two functions occurs for:

$$x=\frac{\pi}{4}(4(1)+1)=\frac{5\pi}{4}$$

Now, observing that:

$$\sin\left(\frac{\pi}{2}\right)>\cos\left(\frac{\pi}{2}\right)$$

$$\sin\left(\frac{3\pi}{2}\right)<\cos\left(\frac{3\pi}{2}\right)$$

We may conclude:

$$\sin(x)>\cos(x)$$ on $$\left[\frac{\pi}{2},\frac{5\pi}{4}\right)$$

$$\cos(x)>\sin(x)$$ on $$\left(\frac{5\pi}{4},\frac{3\pi}{2}\right]$$

And so the area $A$ in question will be given by:

$$A=\int_{\frac{\pi}{2}}^{\frac{5\pi}{4}} \sin(x)-\cos(x)\,dx+\int_{\frac{5\pi}{4}}^{\frac{3\pi}{2}} \cos(x)-\sin(x)\,dx=(1+\sqrt{2})+(\sqrt{2}-1)=2\sqrt{2}$$

Is this question correct? We are given to evaluate:

$$\int_0^2 \left(e^x-e^{-x}\right)^2\,dx$$

$$2\left(\frac{1}{2}\sinh(x)-x\right)$$

$$2\left(\frac{1}{2}\sinh(2\cdot2)-2\right)-2\left(\frac{1}{2}\sinh(2\cdot0)-0\right)$$

$$\sinh(4)-4$$

 

[MarkFL]

Is this question correct? We are given to evaluate:

$$\int_0^2 \left(e^x-e^{-x}\right)^2\,dx$$

$$2\left(\frac{1}{2}\sinh(x)-x\right)$$

$$2\left(\frac{1}{2}\sinh(2\cdot2)-2\right)-2\left(\frac{1}{2}\sinh(2\cdot0)-0\right)$$

$$\sinh(4)-4$$

I have moved this post to its own thread, because we really don't want threads to become a series of one question after another. It is best for forum organization for each question to be in its own thread.

Your result is correct...I would have written essentially the same thing:

$$\int_0^2 \left(e^x-e^{-x}\right)^2\,dx=4\int_0^2 \sinh^2(x)\,dx=2\int_0^2 \cosh(2x)-1\,dx=2\left[\frac{1}{2}\sinh(2x)-x\right]_0^2=2\left(\frac{1}{2}\sinh(4)-2-\frac{1}{2}\sinh(0)+0\right)=\sinh(4)-4$$

 

[Greg]

Integration by parts:

$$\begin{align*}\int\sinh^2(x)\,\text{d}x&=\sinh(x)\cosh(x)-\int\cosh^2(x)\,\text{d}x \\
&=\sinh(x)\cosh(x)-x-\int\sinh^2(x)\,\text{d}x \\
2\int\sinh^2(x)\,\text{d}x&=\sinh(x)\cosh(x)-x \\
\int\sinh^2(x)\,\text{d}x&=\frac12(\sinh(x)\cosh(x)-x)+C\end{align*}$$

 

[HallsofIvy]

Or: [math]\int_0^2 (e^x- e^{-x})^2 dx=\int_0^2 e^{2x}- 2+e^{-2x} dx=\left[\frac{1}{2}e^{2x}-2x- \frac{1}{2}e^{-2x}\right]_0^2= \left(\frac{e^4}{2}- 4- \frac{e^{-4}}{2}\right)[/math]