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We would like to not have that square root there. Luckily, there's a way of turning cosine into cosine squared: the identity $\cos(2x) = 2\cos^2x -1$. If we take $2x=t$, then $\cos t = 2\cos^2(t/2)-1$.

$$\begin{aligned} \int_0^{\pi/2}\sqrt{\cos t+1}\ \mathrm{d}t&=\int_0^{\pi/2}\sqrt{2\cos^2 (t/2)}~\mathrm{d}t = \sqrt{2}\int_0^{\pi/2} |\cos (t/2)|~\mathrm{d}t \\ \text{Over the interval $[0,\frac{\pi}{2}]$, $\cos (t/2) >0$, so we can drop the absolute values.} \\ &= \sqrt{2}\int_0^{\pi/2} \cos (t/2)~\mathrm{d}t = \sqrt{2}\left[2\sin\left(\frac{t}{2}\right)\right]_0^{\pi/2}\\ &=2\sqrt{2}\sin\left(\frac{\pi}{4}\right) = {2} \end{aligned}$$

• Solution 1: Using logarithm rules, $\log\sqrt{x} = \log\left(x^{1/2}\right)=\frac{1}{2}\log x$, so we can simplify: $$\int_1^e \frac{\log\sqrt{x}}{{x}}\mathrm{d}x =\int_1^e \frac{\log{x}}{{2x}}\mathrm{d}x$$

We use the substitution $u=\log x$, $\mathrm{d}u=\frac{1}{x}\mathrm{d}x$:

$$\begin{aligned} \int_1^e \frac{\log{x}}{{2x}}\mathrm{d}x&=\frac{1}{2}\int_1^e \underbrace{\log(x)}_{u}\cdot\underbrace{\frac{1}{x}\,\mathrm{d}x}_{\mathrm{d}u}\\ &=\frac{1}{2}\int_{\log(1)}^{\log(e)}u\,\mathrm{d}u\\ &=\frac{1}{2}\int_{0}^{1}u\,\mathrm{d}u\\ &=\frac{1}{2}\left[ \frac{1}{2}u^2\right]_0^1\\ &=\frac{1}{2}\left[\frac{1}{2}-0\right]=\dfrac{1}{4} \end{aligned}$$

• Solution 2: We use the substitution $u=\log\sqrt{x}$. Then $\dfrac{\mathrm{d}u}{\mathrm{d}x}=\dfrac{1}{\sqrt{x}}\cdot\dfrac{1}{2\sqrt{x}}=\dfrac{1}{2x}$, hence $2\mathrm{d}u=\dfrac{1}{x}\mathrm{d}x$. This fits our integral nicely!

$$\begin{aligned} \int_1^e\frac{\log\sqrt{x}}{x}\mathrm{d}x &= \int_{\log\sqrt{1}}^{\log\sqrt{e}}u\cdot 2\mathrm{d}u \\ &=\Big[u^2\Big]_{0}^{1/2}\\ &=\left(\frac{1}{2}\right)^2-0^2=\frac{1}{4} \end{aligned}$$

$$\int_{0.1}^{0.2} \frac{\tan x}{\log(\cos x)} \mathrm{d}x$$

It might not be immediately obvious how to proceed on this one, so this is another example of an integral where you should not be discouraged by finding methods that don't work. One thing that's worked for us in the past is to use a $u$-substitution with the denominator. With that in mind, let's find the derivative of the denominator. $$\frac{\mathrm{d}}{\mathrm{d}x}\left\{\log(\cos x)\right\}= \frac{1}{\cos x}\cdot (-\sin x) = \frac{-\sin x}{\cos x} = -\tan x$$

So, if we let $u=\log(\cos x$), we see $-\mathrm{d}u=\tan x \mathrm{d}x$, which will work for a substitution.

$$\begin{aligned} \int_{0.1}^{0.2} \frac{\tan x}{\log(\cos x)} \mathrm{d}x&= \int_{\log(\cos(0.1))}^{\log(\cos(0.2))} \frac{-\mathrm{d}u}{u}\\ &=\Big[-\log|u|\Big]_{\log(\cos(0.1))}^{\log(\cos(0.2))} \\ &= -\log|\log(\cos 0.2)|+\log|\log(\cos 0.1)|\\ &={\log\left|\frac{\log(\cos(0.1))}{\log(\cos(0.2))}\right|}\\&= \log\left(\frac{\log(\cos(0.1))}{\log(\cos(0.2))}\right) \end{aligned}$$

Things to notice: the integrand is only defined when $\log(\cos x)$ exists AND is nonzero. So, for instance, it is not defined when $x=0$, because then $\log\cos x = \log1=0$, and we can't divide by zero.

In the final simplification, since $0.1$ and $0.2$ are between 0 and $\pi/2$, the cosine term is positive but less than one, so $\log(\cos 0.1)$ and $\log(\cos 0.2)$ are both negative; then their quotient is positive, so we can drop the absolute value signs.

Using the base change formula, we can also write the final answer as $\log\left(\log_{\cos (0.2)}\cos(0.1)\right)$.

(a) If we expand the integrand, one part of it is quite familiar–a portion of a circle. So, we split the specified integral in two.

$$\int_0^3(x+1)\sqrt{9-x^2} \ \mathrm{d}x = \int_0^3\sqrt{9-x^2} \ \mathrm{d}x +\int_0^3x\sqrt{9-x^2} \ \mathrm{d}x$$

The first piece represents the area above the $x$–axis and below the curve $y=\sqrt{9-x^2}$, i.e. $x^2+y^2=9$, with $0\le x\le 3$. That's the area of one quadrant of a disk of radius $3$. So

$$\int_0^3\sqrt{9-x^2} \ \mathrm{d}x =\frac{1}{4}(\pi \cdot 3^2) = \frac{9}{4}\pi$$

For the second part, we substitute $u=9-x^2$, $\mathrm{d}u= -2x\,\mathrm{d}x$. Note $u(0)=9$ and $u(3)=0$. So,

$$\begin{aligned} \int_0^3x\sqrt{9-x^2} \ \mathrm{d}x = \int_9^0\sqrt{u} \ \frac{\mathrm{d}u}{-2} =-\frac{1}{2}\left[\frac{u^{3/2}}{3/2}\right]_9^0 =-\frac{1}{2}\left[-\frac{27}{3/2}\right] =9 \end{aligned}$$

All together,

$$\int_0^3(x+1)\sqrt{9-x^2} \ \mathrm{d}x = \frac{9}{4}\pi + 9$$

(b) The integrand is of the form $N(x)/D(x)$ with $D(x)$ already factored and $N(x)$ of lower degree. We immediately look for a partial fractions decomposition:

$$\frac{4x+8}{(x-2)(x^2+4)} = \frac{A}{x-2} + \frac{Bx+C}{x^2+4}.$$

Multiplying through by the denominator yields

$$\begin{aligned} 4x+8 &= A(x^2+4) + (Bx+C)(x-2) \tag{$*$} \end{aligned}$$

Setting $x=2$ we find:

$$\begin{aligned} 8 +8 = A(4 + 4) + 0 \implies 16 =8A \implies A = 2 \end{aligned}$$

Substituting $A=2$ in $(*)$ gives

$$\begin{aligned} &&4x+8 &= A(x^2+4) + (Bx+C)(x-2) \\ &\implies & -2x^2+4 x &=(x-2)(Bx + C) \\ &\implies\qquad & (-2x)(x-2)&=(Bx + C)(x-2) \\ & \implies& B& =-2,~C=0 \end{aligned}$$

So we have found that $A=2$, $B=-2$, and $C=0$. Therefore

$$\begin{aligned} \int\frac{4x+8}{(x-2)(x^2+4)}\ \mathrm{d}x &= \int \bigg( \frac{{red}2}{x-2} {red}- \frac{{red}2x}{x^2+4} \bigg) \,\mathrm{d}x \\ &= 2 \log |x-2| - \log (x^2 + 4) + C \end{aligned}$$

Here the second integral was found just by guessing an antiderivative. Alternatively, one could use the substitution $u=x^2+4$, $\mathrm{d}u=2x\,\mathrm{d}x$.

(c) The given integral is improper, but only because of its infinite limits of integration. (The integrand is continuous for all real numbers.) So, we'll have to take two limits. Before we do that, though, let's find the antiderivative. We would like to use the substitution $u=e^x$, $\mathrm{d}u=e^x~\mathrm{d}x$. That is, $\frac{1}{u}~\mathrm{d}u = \mathrm{d}x$.

$$\begin{aligned} \int \frac{1}{e^x+e^{-x}}~\mathrm{d}x&=\int \frac{1}{u(u+\frac{1}{u})}~\mathrm{d}u = \int\frac{1}{u^2+1} = \arctan u+C \\&=\arctan(e^x)+C \\ \text{Now we can deal with the limits of integration.} \\ \int _{ -\infty}^\infty\frac{1}{e^x+e^{-x}}~\mathrm{d}x&= \int _{ -\infty}^0\frac{1}{e^x+e^{-x}}~\mathrm{d}x+ \int _{ 0}^\infty\frac{1}{e^x+e^{-x}}~\mathrm{d}x\\&= \lim_{a \to -\infty}\left[\int_{a}^0\frac{1}{e^x+e^{-x}}~\mathrm{d}x\right]+ \lim_{b \to \infty}\left[\int_{ 0}^b\frac{1}{e^x+e^{-x}}~\mathrm{d}x\right]\\&= \lim_{a \to -\infty}\big[\arctan(e^x)\big]_a^0+ \lim_{b \to \infty}\big[\arctan(e^x)\big]_0^b\\&= \lim_{a \to -\infty}\big[\arctan(e^0)-\arctan(e^a)\big]+ \lim_{b \to \infty}\big[\arctan(e^b)-\arctan(e^0)\big]\\&= \lim_{a \to -\infty}\big[-\arctan(e^a)\big]+ \lim_{b \to \infty}\big[\arctan(e^b)\big]\\ &=-\arctan(0)+\frac{\pi}{2} = \frac{\pi}{2} \end{aligned}$$

It's not immediately clear where to start, but a common method we've seen is to use the denominator in a $u$-substitution, especially when square roots are involved.

Let $u = \sqrt{1-x}$, $\mathrm{d}u = -\frac{1}{2\sqrt{1-x}}\mathrm{d}x$. Then $u^2=1-x$, so $x = 1-u^2$.

$$\begin{aligned} \int \sqrt{\frac{x}{1-x}}~\mathrm{d}x&=2\int \frac{\sqrt{x}}{2\sqrt{1-x}}~\mathrm{d}x =- 2\int\sqrt{1-u^2}~\mathrm{d}u \\ \text{Now we're back in familiar territory. Let $u=\sin \theta$, $\mathrm{d}u=\cos \theta~\mathrm{d}\theta$.} \\ &=-2\int\sqrt{1-\sin^2 \theta }\cos\theta~\mathrm{d}\theta\\ &=-2\int\cos^2\theta~\mathrm{d}\theta\\ &=-\int \big(1+\cos(2\theta)\big)~\mathrm{d}\theta\\ &=-\theta - \frac{1}{2}\sin(2\theta)+C\\ &=-\theta - \sin\theta\cos\theta+C\\ &=-\arcsin u - u\sqrt{1-u^2}+C \tag{$*$}\\ \smash{  }\,&=-\arcsin (\sqrt{1-x})-\sqrt{1-x}\sqrt{x}+C \end{aligned}$$

In ($*$), to convert from $\theta$ to $u$, our substitution $u=\sin\theta$ tells us $\theta = \arcsin u$. To find $\cos \theta$, we can either trace our work backwards to see that we already simplified $\sqrt{1-u^2}$ into $\cos\theta$, or we can draw a right triangle with angle $\theta$ and $\sin \theta=u$, then use the Pythagorean theorem to find the length of the adjacent side of the triangle and $\cos\theta$.

Let's use the substitution $u=e^x$. There are a few reasons to think this is a good choice. It's an “inside function," in that if we let $f(x)=e^x$, then $f(e^x)=e^{e^x}$, which is a piece of our integrand. Also its derivative, $e^x$, is multiplied by the rest of the integrand, since $e^{2x}=e^x\cdot e^x$ .

Let $u=e^x$, $\mathrm{d}u = e^x\mathrm{d}x$. When $x=0$, $u=1$, and when $x=1$, $u=e$.

$$\begin{aligned} \displaystyle\int_0^1e^{2x}e^{e^x}\mathrm{d}x &= \int_0^1e^{x}e^{e^x}e^x\mathrm{d}x =\int_1^e ue^u~\mathrm{d}u \\ \text{This is more familiar. We use integration by parts with $\mathrm{d}v = e^u~\mathrm{d}u$, $v=e^u$. Conveniently, the ``$u$" we brought in with the substitution is what we want to use for the ``$u$" in integration by parts, so we don't have to change the names of our variables.} \\ &=\big[ue^u\big]_1^e - \int_1^e e^u~\mathrm{d}u\\ &=e\cdot e^e-e-e^e+e=e^e(e-1) \end{aligned}$$

The substitution $u=x+1$ looks promising at first, but doesn't result in something easily integrable. We can't use partial fractions because our integration isn't rational. This doesn't look like something from the trig-substitution family. So, let's think about integration by parts. There's a lot of different ways we could break up the integrand into two parts. For example, we could view it as $\Big(\frac{x}{(x+1)^2}\Big)\Big(e^x\Big)$, or we could view it as $\Big(\frac{x}{x+1}\Big)\Big(\frac{e^x}{x+1}\Big)$. After some trial and error, we settle on $u=xe^x$ and $\mathrm{d}v=(x+1)^{-2}~\mathrm{d}x$. Then $\mathrm{d}u=e^x(x+1)$ and $v=\frac{-1}{x+1}$.

$$\begin{aligned} \int\frac{xe^x}{(x+1)^2}~\mathrm{d}x&=-\frac{xe^x}{x+1} +\int \frac{e^x(x+1)}{x+1}\mathrm{d}x \\&=-\frac{xe^x}{x+1} +\int e^x\mathrm{d}x\\ &=-\frac{xe^x}{x+1}+e^x+C \\&= \frac{e^x}{x+1}+C \end{aligned}$$