# MA 1505 Tutorial 2: Integration

L.Hospital Rule: if the ratio is $\infty/\infty$ or 0/0, then we can use the L.Hospital Rule to calculate the limit. Precisely, the L.Hospital Rule is

$\lim_{x\rightarrow a} \frac{f(x)}{g(x)}=\lim_{x\rightarrow a} \frac{f^{'}(x)}{g^{'}(x)},$

where a is a finite real number or infinity.

If f(x) is a continuous function, then $F(x)= \int_{a}^{x} f(t) dt$ is a differentiable function and its derivative $F^{'}(x)=f(x)$.

General Leibniz Integration Rule.

$\frac{d}{d\theta} ( \int_{a(\theta)}^{b(\theta)} f(x,\theta)dx)$

$= \int_{a(\theta)}^{b(\theta)} f_{\theta}(x,\theta) dx + f(b(\theta), \theta)\cdot b^{'}(\theta) - f(a(\theta),\theta) \cdot a^{'}(\theta)$

Question 1. Calculate the value $S=\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\sin x+\cos x}dx$.

Solution.

$S=\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\sin x +\cos x }dx$

$= \int_{0}^{\frac{\pi}{2}} \frac{\cos(\frac{\pi}{2}-t)}{\sin(\frac{\pi}{2}-t) + \cos( \frac{\pi}{2}-t)} dt$

$= \int_{0}^{\frac{\pi}{2}} \frac{\sin t}{\cos t+\sin t}dt$

Therefore

$2S=S+S$

$= \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\sin x+\cos x} + \frac{\sin x}{\cos x+\sin x} dx$

$= \int_{0}^{\frac{\pi}{2}} 1 dx= \frac{\pi}{2}$

Hence, $S=\frac{\pi}{4}$.

Question 2. Calculate the value $S=\int_{0}^{\frac{\pi}{4}} \ln(1+\tan{x}) dx$.

Solution.

$S=\int_{0}^{\frac{\pi}{4}} \ln(1+\tan x) dx$

$= \int_{0}^{\frac{\pi}{4}} \ln(1+\tan(\frac{\pi}{4}-x)) dx$

$= \int_{0}^{\frac{\pi}{4}} \ln(\frac{2}{1+\tan x})dx$

$= \frac{\pi \ln2}{4} - \int_{0}^{\frac{\pi}{4}} \ln(1+\tan x) dx$

$= \frac{\pi \ln2}{4} - S$

Therefore, $S=\frac{\pi \ln2}{8}$.

Question 3.

$S=\int_{0}^{1} \frac{x^{4}}{x^{4}+(1-x)^{4}}dx$

Solution.

$S=\int_{0}^{1} \frac{x^{4}}{x^{4}+(1-x)^{4}}dx$

$= \int_{0}^{1} \frac{(1-x)^{4}}{x^{4}+(1-x)^{4}}dx$

$= 1- \int_{0}^{1} \frac{x^{4}}{x^{4}+(1-x)^{4}}dx$

Therefore, $S=0.5.$

Question 4.

$\lim_{x\rightarrow 0} \frac{\sin x}{x}=1.$

$\lim_{x\rightarrow \infty} x\tan\frac{1}{x} = \lim_{y\rightarrow 0}\frac{\tan y}{y}=1,$ where y=1/x.

$\lim_{x\rightarrow \infty} \frac{\ln x}{x^{a}}=0,$ where a>0.