Michael Bowen's VC Course Pages

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Math V21B Section 7.1

Introduction

Please try this problem on your own before looking at my solution. There may be other ways to solve this; I make no claim that this method is the quickest, prettiest, or most efficient. Also, I did not look at the student solutions manual, so this approach may or may not match what they did.

Problem 53

The original problem is to prove that $$\displaystyle{\int {{{\tan }^n}x\;dx} = \frac{{{{\tan }^{n - 1}}x}}{{n - 1}} - \int {{{\tan }^{n - 2}}x\;dx}}$$

where $n$ is a natural number other than 1. This appears to be a straight integration by parts, similar to problem #51. However, when it comes to trying integration by parts, any reasonable choices of $u$ and $dv$ do not give satisfactory results, as you may have discovered. This suggests that the problem is a bit deeper than it appears at first glance. Restricting ourselves to expressions involving only powers of the tangent function is not serving us well, so let us consider writing this function in terms of sines and cosines instead. The left-hand side of the equation may be rewritten as $$\displaystyle{\int {\frac{{{{\sin }^n}x}}{{{{\cos }^n}x}}dx}}$$ after which we might be inclined to set $u$ equal to either the numerator ${{\sin }^n}x$ or the denominator ${{\cos }^n}x$). Experimentation reveals, however, that neither of these choices is helpful, either. However, if we set $u = {\sin ^{n - 1}}x$, helpful things do begin to happen. This choice forces $du = \left( {n - 1} \right){\sin ^{n - 2}}x\cos x\;dx$ and $\displaystyle{dv = \frac{{\sin x}}{{{{\cos }^n}x}}dx}$.

We need to integrate the last expression to obtain $v$; setting up a $w$-substitution with $w = \cos x$ and $dw = - \sin x\;dx$ allows us to determine that $$\displaystyle{v = \int {dv} = \int {\frac{{ - dw}}{{{w^n}}}} = - \int {{w^{ - n}}\;dw}}$$ and since we are assured by hypothesis (above) that $n \ne 1$, we conclude that $$\displaystyle{v = - \frac{w^{ - n + 1}}{ - n + 1} = \frac{1}{n - 1} \cdot \frac{1}{w^{n - 1}} = \frac{1}{n - 1} \cdot \frac{1}{{{\cos }^{n - 1}}x}}$$ (in the above equation, we first apply the power rule to the last integral in the second equation above, then distribute the negative sign preceding the first fraction above through the denominator and move the $w$ to the denominator while multiplying its exponent by $-1$, and finally back-substitute $\cos x$ for $w$). The resulting expression is guaranteed to have a positive exponent on the cosine function, as $n\ge 2$ by hypothesis.

Applying the principle of integration by parts to the rewritten integral yields $$\int {{{\tan }^n}x\;dx} = uv - \int {v\;du}$$ $$\displaystyle{\int {{{\tan }^n}x\;dx} = {\sin ^{n - 1}}x \cdot \frac{1}{n - 1} \cdot \frac{1}{{{\cos }^{n - 1}}x} - \int {\frac{1}{n - 1} \cdot \frac{1}{{{\cos }^{n - 1}}x} \cdot \left( {n - 1} \right)\cdot {{\sin }^{n - 2}}x\cos x\;dx}}$$ We simplify the right side of the equation by consolidating the sine and cosine factors in the first term. In the second term (the remaining integrand), we initially consolidate the powers of all the cosine factors and cancel the constants $\left(n - 1\right)$, leading to $$\displaystyle{\int {{{\tan }^n}x\;dx} = \frac{1}{{n - 1}} \cdot \frac{{\sin ^{n - 1}}x}{{{{\cos }^{n - 1}}x}} - \int {{{\cos }^{1 - \left(n - 1\right)}}x \cdot {{\sin }^{n - 2}}x\;dx}}$$

The next step of the simplification of the right side of the equation involves simplifying the matching sine and cosine powers in the first term into the same power of the tangent, simplifying the powers of the cosine in the integrand, and moving that cosine to the denominator by negating the signs on its exponents: $$\displaystyle{\int {{{\tan }^n}x\;dx} = \frac{1}{{n - 1}} \cdot {{{{\tan }^{n - 1}}x}} - \int {\frac{1}{{{{\cos }^{n - 2}}x}} \cdot {{\sin }^{n - 2}}x\;dx}}$$ We consolidate the factors in both terms on the right side of the equation to complete the proof: $$\displaystyle{\boxed{\int {{{\tan }^n}x\;dx} = \frac{{{\tan }^{n - 1}}x}{n - 1} - \int {{{\tan }^{n - 2}}x\;dx}}}$$

As an aside, this result has practical value, as it permits us to use recursion (a mathematical word that means "using previously-obtained results repetitively as stepping-stones to obtain new results") to find the antiderivatives of all positive-integer powers of the tangent function. From past work, we know that $$\displaystyle{\int {\tan x\;dx} = \underline{\underline{ \ln \left| {\sec x} \right| + C}}}$$ Plugging $n=2$ into the formula whose proof we just obtained yields $$\displaystyle{\int {{{\tan }^2}x\;dx} = \frac{{{\tan }^{2 - 1}}x}{2 - 1} - \int {{{\tan }^{2 - 2}}x\;dx} = \frac{\tan x}{1} - \int {{{\left( {\tan x} \right)}^0}\;dx} = \tan x - \int {dx} = \underline{\underline {\tan x - x + C}}}$$ Continuing with successively larger values of $n$ (and re-using results from previous $n$ values to evaluate the later antiderivatives) yields $$\displaystyle{\int {{{\tan }^3}x\;dx} = \frac{{{\tan }^{3 - 1}}x}{3 - 1} - \int {{{\tan }^{3 - 2}}x\;dx} = \frac{{{\tan }^2}x}{2} - \int {\tan x\;dx} = \underline{\underline {\frac{{{\tan }^2}x}{2} - \ln \left| {\sec x} \right| + C}}}$$ $$\displaystyle{\int {{{\tan }^4}x\;dx} = \frac{{{\tan }^{4 - 1}}x}{4 - 1} - \int {{{\tan }^{4 - 2}}x\;dx} = \frac{{{\tan }^3}x}{3} - \int {{{\tan }^2}x\;dx} = \underline{\underline {\frac{{{\tan }^3}x}{3} - \tan x + x + C}}}$$ $$\displaystyle{\int {{{\tan }^5}x\;dx} = \frac{{{\tan }^{5 - 1}}x}{5 - 1} - \int {{{\tan }^{5 - 2}}x\;dx} = \frac{{{\tan }^4}x}{4} - \int {{{\tan }^3}x\;dx} = \underline{\underline {\frac{{{\tan }^4}x}{4} - \frac{{{\tan }^2}x}{2} + \ln \left| {\sec x} \right| + C}}}$$ $$\displaystyle{\int {{{\tan }^6}x\;dx} = \frac{{{\tan }^{6 - 1}}x}{6 - 1} - \int {{{\tan }^{6 - 2}}x\;dx} = \frac{{{\tan }^5}x}{5} - \int {{{\tan }^4}x\;dx} = \underline{\underline {\frac{{{\tan }^5}x}{5} - \frac{{{\tan }^3}x}{3} + \tan x - x + C}}}$$ etc., giving some of the same results as section 7.2 but with a different technique.