Example integral question that contains more than 2 Integral Techniques

Question

For example, the integral $$ \int \sqrt{1-x^{2}} \: dx $$ Can be solved using trigonometric substitution and partial integration. $$ \int \sqrt{1-x^{2}} \: dx = \int \cos^{2}(t) dt $$ with $x=\sin(t)$ substitution. Then the latter can be solved using partial integration : $$\int \cos^{2}(t) dt = \sin(t)\cos(t) + \int \sin^{2}(t) dt $$ and so on..

Another one, which is a good one, is

$$ \int \ln(x^{2} - 2x +5) \: dx$$ This can be solved using combination of Partial integration, then subs $u=x-1$, and then Trig.subs $u= 2 \tan(t)$. In that order.

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My question is : Is there one or two great sample problems such that we must use all these integration techniques : usual subs, partial integration, trig.subs, and also algebraic manipulation, in undergraduate level? First year Calculus

This is for an assignment, so that perhaps one problem is enough to represent all techniques and material.

I would appreciate the inputs, thanks.

Answer 1

Famously, expressing $\int\sec^{n+2}x dx$ in terms of $\int\sec^{n}x dx$ requires parts and algebraic manipulation. So let's start with the challenge of integrating $\sec^3 x$ and make it harder, as per your request, as $\int\cos x\sec^3 \sin x dx$. And we can add a "usual" substitution, viz. the calculation $f(y):=\int_0^y x\cos (1+x^2)\sec^3 \sin (1+x^2) dx$. And now to add in a use of differentiation under the integral sign, we pose the problem $\int_0^\infty z e^{-zf(y)}dz$.

Answer 2

A basic integral by parts is

$$\int \log x \: dx=x\log x-\int 1 \,dx$$

and by substitution (by polar coordinates)

$$\int e^{x^2+y^2} \,dx\,dy$$

another basic integral by parts is

$$\int \arctan x \: dx=x\arctan x-\int \frac{x}{1+x^2} \,dx$$

Answer 3

I found one, it is from a book.

$$ \int \ln(x^{2} - 2x +5) \: dx$$ This can be solved using combination of Partial integration, then subs $u=x-1$, and then Trig.subs $u= 2 \tan(t)$. In that order.

But i am still open for another answer. Thanks.

Answer 4

I will find the integral $\displaystyle{\int \frac{dx}{1 - x^2}}$ using five different methods. I hope this is the kind of thing you are after.

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Method 1 - Using a partial fraction decomposition

This is perhaps the easiest method of the five. Since $$\frac{1}{1 - x^2} = \frac{1}{(1 - x)(1 + x)} = \frac{1}{2} \left [\frac{1}{1 - x} + \frac{1}{1 + x} \right ],$$ we have $$\int \frac{dx}{1 - x^2} = \frac{1}{2} \int \left [\frac{1}{1 - x} + \frac{1}{1 + x} \right ] \, dx = \frac{1}{2} \ln \left |\frac{1 + x}{1 - x} \right | + C.$$

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Method 2 - Using a trigonometric substitution

This is also a relatively easy method to use.

Let $x = \sin \theta, dx = \cos \theta \, d\theta$. Thus \begin{align*} \int \frac{dx}{1 - x^2} &= \int \frac{\cos \theta}{1 - \sin^2 \theta} \, d\theta\\ &= \int \frac{\cos \theta}{\cos^2 \theta} \, d\theta\\ &= \int \sec \theta \, d\theta\\ &= \int \sec \theta \cdot \frac{\sec \theta + \tan \theta}{\sec \theta + \tan \theta} \, d\theta\\ &= \int \frac{\sec^2 \theta + \sec \theta \tan \theta}{\sec \theta + \tan \theta} \, d\theta\\ &= \ln |\sec \theta + \tan \theta| + C\\ &= \ln \left |\frac{1 + x}{\sqrt{1 - x^2}} \right | + C\\ &= \ln \sqrt{\frac{1 + x}{1 - x}} + C\\ &= \frac{1}{2} \ln \left |\frac{1 + x}{1 - x} \right | + C. \end{align*}

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Method 3 - Using a hyperbolic substitution

Is very easy provided you are comfortable working with hyperbolic functions.

Let $x = \tanh u, dx = \text{sech}^2 u \, du$. Thus \begin{align*} \int \frac{dx}{1 - x^2} &= \int \frac{\text{sech}^2 u}{1 - \tanh^2 u} \, du\\ &= \int \frac{\text{sech}^2 u}{\text{sech}^2 u} \, du\\ &= \int du\\ &= u + C\\ &= \tanh^{-1} x + C\\ &= \frac{1}{2} \ln \left |\frac{1 + x}{1 - x} \right | + C, \end{align*} where in the last line the well-known result for the inverse hyperbolic tangent function expressed in terms of the natural logarithmic function has been used.

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Method 4 - Using an algebraic substitution

If you are fortunate enough to see it, the integral can also be found using a self-similar substitution of $u = \dfrac{1 - x}{1 + x}$.

Here we see that $x = \dfrac{1 - u}{1 + u}$ such that $dx = -\dfrac{2}{(1 + u)^2} \, du$. Under this substitution we see that $$1 - x^2 = \frac{4u}{(1 + u)^2}.$$ Thus \begin{align*} \int \frac{dx}{1 - x^2} &= -\int \frac{(1 + u)^2}{4u} \cdot \frac{2}{(1 + u)^2} \, du\\ &= -\frac{1}{2} \int \frac{du}{u}\\ &= -\frac{1}{2} \ln |u| + C\\ &= -\frac{1}{2} \ln \left |\frac{1 - x}{1 + x} \right | + C\\ &= \frac{1}{2} \ln \left |\frac{1 + x}{1 - x} \right | + C. \end{align*}

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Method 5 - Using integration by parts

This is by far the most circuitous method of them all.

Integrating by parts we have \begin{align*} \int \frac{dx}{1 - x^2} &= \frac{x}{1 - x^2} - 2 \int \frac{x^2}{(1 - x^2)^2} \, dx\\ &= \frac{x}{1 - x^2} + 2 \int \frac{(1 - x^2) - 1}{(1 - x^2)^2} \, dx\\ &= \frac{x}{1 - x^2} + 2 \int \frac{dx}{1 - x^2} - 2 \int \frac{dx}{(1 - x^2)^2}\\ \Rightarrow - \int \frac{dx}{1 - x^2} &= \frac{x}{1 - x^2} - 2 \int \frac{dx}{(1 - x^2)^2}, \end{align*} or $$\int \frac{dx}{1 - x^2} = -\frac{x}{1 - x^2} + 2 \int \frac{dx}{(1 - x^2)^2}.\tag1$$ To find the integral appearing on the right, let $x = \sin \theta, dx = \cos \theta \, d\theta$. Thus \begin{align*} \int \frac{dx}{(1 - x^2)^2} &= \int \frac{\cos \theta}{\cos^3 \theta} \, d\theta\\ &= \int \sec^3 \theta \, d\theta\\ &= \int \sec^2 \theta \cdot \sec \theta \, d\theta\\ &= \tan \theta \sec \theta - \int \sec \theta \tan^2 \theta \, d\theta \quad \text{(by parts)}\\ &= \tan \theta \sec \theta - \int \sec \theta (\sec^2 \theta - 1) \, d\theta\\ &= \tan \theta \sec \theta + \int \sec \theta \, d\theta - \int \sec^3 \theta \, d\theta. \quad (*) \end{align*} Since $$\int \sec \theta \, d\theta = \ln |\sec \theta + \tan \theta| + C \quad \text{(was found in Method 2)},$$ and as the integral for $\sec^3 \theta$ has reappeared, we have \begin{align*} \int \frac{dx}{(1 - x^2)^2} &= \frac{1}{2} \tan \theta \sec \theta + \frac{1}{2} \ln |\sec \theta + \tan \theta| + C\\ &= \frac{1}{2} \frac{x}{1 - x^2} + \frac{1}{4} \ln \left |\frac{1 + x}{1 - x} \right | + C. \end{align*} Thus (1) becomes $$\int \frac{dx}{1 - x^2} = -\frac{x}{1 - x^2} + \frac{x}{1 - x^2} + \frac{1}{2} \ln \left |\frac{1 + x}{1 - x} \right | + C = \frac{1}{2} \ln \left |\frac{1 + x}{1 - x} \right | + C,$$ as expected.