integrate((cos(2*t))^4,t);
integrate(cos(3*t)*sin(8*t),t);
integrate(sin(8*x)*sin(x),x,1,3);
float(integrate(sin(8*x)*sin(x),x,1,3));
integrate(cot(10*x)*(csc(10*x))^4,x);
| Form | Looks Like | Substitution | Limit Assumptions |
|---|---|---|---|
| $$\sqrt{b^2x^2-a^2}$$ | $$\sec^2(\theta)-1=\tan^2(\theta)$$ | $$x=\frac{a}{b}\sec(\theta)$$ | $$0\le\theta<\frac{\pi}{2},\,\frac{\pi}{2}<\theta\le \pi$$ |
| $$\sqrt{a^2-b^2x^2}$$ | $$1-\sin^2(\theta)=\cos^2(\theta)$$ | $$x=\frac{a}{b}\sin(\theta)$$ | $$-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$$ |
| $$\sqrt{a^2+b^2x^2}$$ | $$\tan^2(\theta)+1=\sec^2(\theta)$$ | $$x=\frac{a}{b}\tan(\theta)$$ | $$-\frac{\pi}{2}<\theta<\frac{\pi}{2}$$ |
Use a trig substitution to eliminate the root.
f1(x):=sqrt(4-9*x^2)$
f1((2/3)*sec(theta));
factor(f1((2/3)*sec(theta)));
Notice that $2\sqrt{1-\sec^2(\theta)}=2\sqrt{\tan^2(\theta)}=2\tan(\theta)$. Thus, by letting $x=\frac{2}{3}\sec(\theta)$ we have $$\sqrt{4-9x^2}=2\tan(\theta)$$
f2(x):=sqrt(3+25*x^2)$
f2((sqrt(3)/5)*tan(theta));
factor(f2((sqrt(3)/5)*tan(theta)));
Notice that $\sqrt3\sqrt{\tan^2(\theta)+1}=\sqrt3\sqrt{\sec^2(\theta)}=\sqrt3\sec(\theta)$. thus, by letting $x=\frac{\sqrt3}{5}\tan(\theta)$ we have $$\sqrt{3+25x^2}=\sqrt3\sec(\theta)$$
f3(x):=(7*x^2-3)^(5/2)$
f3((sqrt(3)/sqrt(7)*sec(theta)));
factor(f3((sqrt(3)/sqrt(7)*sec(theta))));
3^(5/2)*((tan(theta))^2)^(5/2);
Notice that $3^{5/2}\cdot (\sec^2(\theta)-1)^{5/2} = 3^{5/2}\cdot tan^5(\theta)$. Thus, by letting $x=\frac{\sqrt3}{\sqrt7}\sec(\theta)$ we have $$\left(7x^2-3\right)^{5/2}=3^{5/2}\tan^5(\theta)$$
f4(x):=sqrt((x+3)^2-100)$
f4(10*sec(theta)-3);
factor(f4(10*sec(theta)-3));
Notice that $10\sqrt{\sec^2(\theta)-1}=10\tan(\theta)$. Thus, by letting $x=10\sec(\theta)-3$ we have $$\sqrt{(x+3)^2-100}=10\tan(\theta)$$
f5(x):=sqrt(4*(9*x-5)^2+1)$
solve((1/2)*tan(theta)=9*x-5,x);
f5((tan(theta)+10)/18);
factor(f5((tan(theta)+10)/18));
First, solve the equation: $$\frac12\tan(\theta)=9x-5$$ to get $x=\frac{\tan(\theta)+10}{18}$. Then we notice that $\sqrt{\tan^2(\theta)+1}=\sec(\theta)$. Thus, with $x=\frac{\tan(\theta)+10}{18}$, we have, $$\sqrt{4(9x-5)^2+1}=\sec(\theta)$$
Use a trig substitution to evaluat ethe given integral. $$\int \frac{\sqrt{x^2+16}}{x^4}\, dx$$
integrate(sqrt(x^2+16)/x^4,x);
Top(x):=sqrt(x^2+16)$
Top(4*tan(theta));
factor(Top(4*tan(theta)));
Notice that when $x=4\tan(\theta)$, we have $$\sqrt{x^2+16}=4\sec(\theta)$$
(4*sec(theta))/((4*tan(theta))^4);
trigreduce((4*sec(theta))/((4*tan(theta))^4));
trigreduce((cot(x))^2+1);
So we have:
$$\int \frac{\sqrt{x^2+16}}{x^4}\, dx= \int\frac{\sec(\theta)}{64\tan^4(\theta)}\,dx$$Recall that $\csc^2(\xi)=\cot^2(\xi)+1$. That is,
$$\cot^3(\theta)\cdot \csc(\theta)=\left(\cot(\theta)\csc(\theta)\right)\cdot \cot^2(\theta)$$$$\left(\cot(\theta)\csc(\theta)\right)\cdot \cot^2(\theta) = \left(\cot(\theta)\csc(\theta)\right)\cdot \left(\csc^2(\theta)-1\right)$$$$\int\frac{\sec(\theta)}{64\tan^4(\theta)}\,dx = \frac{1}{64} \int \left(\cot(\theta)\csc(\theta)\right)\cdot \left(\csc^2(\theta)-1\right)\,d\theta$$Let $u=\csc(\theta)$ and $du=\cot(\theta)\csc(\theta)$, then
$$\frac{1}{64} \int \left(\cot(\theta)\csc(\theta)\right)\cdot \left(\csc^2(\theta)-1\right)\,d\theta= \frac{1}{64}\int\left(u^2-1\right)\,du$$IF9(u):=(u*(u^2-3)/192);
IF9(csc(theta));
| Factor in Denominator | Term in Partial Fraction Decomposition |
|---|---|
| $$ax+b$$ | $$\dfrac{A}{ax+b}$$ |
| $$(ax+b)^k$$ | $$\dfrac{A_1}{ax+b}+\dfrac{A_2}{(ax+b)^2}+\cdots+\dfrac{A_k}{(ax+b)^k},\,k=1,2,3,...$$ |
| $$(ax^2+bx+c)^k$$ | $$\dfrac{A_1x+B_1}{ax^2+bx+c}+\dfrac{A_2x+B_2}{(ax^2+bx+c)^2}+\cdots+\dfrac{A_kx+B_k}{(ax^2+bx+c)^k},\,k=1,2,3,...$$ |
Note for number 3: the integral $\int_{-\infty}^{\infty}f(x)\,dx$ is convergent if and only if $\int_{-\infty}^cf(x)\,dx$ and $\int_c^{\infty}f(x)\,dx$ are both convergent.
If $a>0$ then $$\int_a^{\infty}\dfrac{1}{x^p}\,dx$$ is convergent if $p>1$ and divergent if $p\le1$.
If $f(x)\ge g(x)\ge 0$ on the interval $[a,\infty)$, then
Suppose ${\displaystyle \lim_{x\to a}|f(x)|=\lim_{x\to a}|g(x)|=\infty}$ or $\displaystyle \lim_{x\to a}f(x)=\lim_{x\to a}g(x)=0$. If ${\displaystyle \lim_{x\to a}\dfrac{f'(x)}{g'(x)}}$ exists in either the finite or infinite sense, then $$ \lim_{x\to a}\dfrac{f(x)}{g(x)}=\lim_{x\to a}\dfrac{f'(x)}{g'(x)}. $$
The average value of a function $f(x)$ over the interval $[a,b]$ is given by $$f_{ave}=dfrac{1}{b-a}\int_a^bf(x)\,dx$$
If $f(x)$ is continuous function on $[a,b]$ then there is a number $c\in[a,b]$ such that $$\int_a^bf(x)\,dx=f(c)(b-a)$$
If $f'$ is continuous on $[a,b]$, then the length (arc length) of the curve $y=f(x)$ from the point $A=(a,f(a))$ to the point $B=(b,f(b))$ is the value of the integral $$L=\int_a^b\sqrt{1+\left[f'(x)\right]^2}\,dx=\int_a^b\sqrt{1+\left[ \frac{dy}{dx} \right]^2}$$
Let $A(x)=\pi(R(x))^2$ where $R(x)$ is a function of the radius of the circle over the domain $x$. The Volume of the object is then $$V=\int_a^bA(x)\,dx$$
The region between the curve $y=\cos(x)$,$\frac{\pi}{4}\le x\le \frac{3\pi}{4}$, and the x-axis is revolved about the x-axis to generate a solid. Find its volume.
First, look at the graph:
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