## 170- Quiz 11

The 11th quiz is here.

Solutions follow.

Problem 1. Find the derivative of $\displaystyle G(x)=\int_1^{5x^2}(t-\sin(t))\,dt$.

There are two ways to proceed. We can solve this problem by direct calculation: First we compute an antiderivative $H(t)$ of $t-\sin t$, then use it to compute $G(x)=H(5x^2)-H(1)$, and finally use this expression to compute $G'(x)$.

We have $\displaystyle \int (t-\sin t)\,dt=\frac{t^2}2+\cos t+C$, so $\displaystyle G(x)=\left.\frac{t^2}2+\cos t\right|_1^{5x^2}=\frac{(5x^2)^2}2+\cos(5x^2)-\frac12-\cos1$, and

$\displaystyle G'(x)=\frac{2(5x^2)10x}2-\sin(5x^2)10x$ $\displaystyle =(5x^2-\sin(5x^2))10x$.

The second method uses instead the fundamental theorem of calculus directly, and avoids having to compute any antiderivatives: Suppose $f$ is a continuous function on the interval ${}[a,b]$ and that $H$ is an antiderivative of $f$. Recall that this simply means that $H'(t)=f(t)$ for $a.

Suppose $g$ is a differentiable function, and $G$ is defined by $\displaystyle G(x)=\int_a^{g(x)}f(t)\,dt$. By the fundamental theorem of calculus, this means that $\displaystyle G(x)=H(g(x))-H(a)$, and therefore

$\displaystyle G'(x)=H'(g(x))\cdot g'(x)=f(g(x))\cdot g'(x),$

using the chain rule. In the particular case that we are interested in, $a=1$, $g(x)=5x^2$, and $f(t)=t-\sin t$, so

$\displaystyle G'(x)=(5x^2-\sin(5x^2))10x,$

as before.

Problem 2. Find $\displaystyle \int_0^{\sqrt \pi/2} x\sec^2(x^2)\tan(x^2)\,dx$.

We use the substitution method to simplify the integrand. Although this can be done directly, I will do it in two stages for clarity. First, the most obvious difficulty with the expression at hand is the $x^2$ that appears as the argument in the trigonometric functions. This suggests to first try the substitution

$t=x^2$.

Note that $dt=2x\,dx$, and the integrand already contains a factor of $x$, which is a good sign. To rewrite the integral in terms of $t$, we note that if $x=0$, then $t=0^2=0$, while if $x=\sqrt\pi/2$, then $t=(\sqrt\pi/2)^2=\pi/4$, so the integral becomes

$\displaystyle \int_0^{\pi/4}\frac12\sec^2(t)\tan(t)\,dt$.

This expression is still complicated. There are two ways to proceed now. We may use that $(\tan t)'=\sec^2 t$. This suggests the substitution $u=\tan(t)$. Then $du=\sec^2(t)\,dt$. When $t=0$, we have $u=\tan(0)=0$, while when $t=\pi/4$, we have $u=\tan(\pi/4)=1$, so the integral becomes

$\displaystyle \int_0^1 \frac12u\,du=\left.\frac{u^2}4\right|_0^1=\frac14$.

The second way to proceed is to use that $(\sec t)'=(\sec t)(\tan t)$. This suggests the substitution $w=\sec(t)$. Then $dw=\sec(t)\tan(t)\,dt$. When $t=0$, we have $w=\sec(0)=1$, while when $t=\pi/4$, we have $w=\sec(\pi/4)=\sqrt2$, so the integral becomes

$\displaystyle \int_1^{\sqrt 2}\frac12w\,dw=\left.\frac{w^2}4\right|_1^{\sqrt2}=\frac24-\frac14=\frac14$,

as before.

I saw a slightly different presentation of the last approach in some of the solutions: Use the substitution $y=\cos(x^2)$. Then $dy=-2x\sin(x^2)\,dx$. If $x=0$, then $y=1$, while if $x=\sqrt\pi/2$, then $y=\cos(\pi/4)=1/\sqrt2$. Then the integral becomes

$\displaystyle \int_1^{1/\sqrt2}-\frac12\frac1{y^3}\,dy=\left.\frac{y^{-2}}4\right|_1^{1/\sqrt2}=\frac24-\frac14=\frac14$.