Integral puzzles with solutions

My source is usually integration puzzles on exams and in maths societies.

1. $\int_{0}^{\frac{π}{4}} (tan x)^{\frac{1}{3}} d x$

Answer

(1)

\int_{0}^{\frac{π}{4}} (tan x)^{\frac{1}{3}} d x = = \int_{0}^{1} \frac{t ^{\frac{1}{3}}}{1 + t ^{2}} d t \int_{0}^{1} \frac{y}{1 + y ^{3}} d y

And

(2)

\frac{y}{1 + y ^{3}} = \frac{y}{( y + 1 ) ( y ^{2} - y + 1 )} = = \frac{1}{3} (\frac{y + 1}{y ^{2} - y + 1} - \frac{1}{y + 1}) \frac{1}{3} (\frac{y + 1}{( y - \frac{1}{2} ) ^{2} + \frac{3}{4}} - \frac{1}{y + 1})

(3)

\int_{0}^{1} \frac{y}{1 + y ^{3}} d y = = = = = \int_{0}^{1} \frac{1}{3} \frac{y + 1}{( y - \frac{1}{2} ) ^{2} + \frac{3}{4}} d y - \int_{0}^{1} \frac{1}{3} \frac{1}{y + 1} d y \int_{- \frac{1}{2}}^{\frac{1}{2}} \frac{1}{3} \frac{x + \frac{3}{2}}{x ^{2} + \frac{3}{4}} d x - \frac{1}{3} ln 2 \int_{- \frac{1}{2}}^{\frac{1}{2}} \frac{1}{2} \frac{1}{x ^{2} + \frac{3}{4}} d x + \int_{- \frac{1}{2}}^{\frac{1}{2}} \frac{1}{3} \frac{x}{x ^{2} + \frac{3}{4}} d x - \frac{1}{3} ln 2 \frac{1}{2} [\frac{2}{3} arctan (\frac{2 x}{3})]_{- \frac{1}{2}}^{\frac{1}{2}} - \frac{1}{3} ln 2 \frac{π}{3 3} - \frac{1}{3} ln 2

Answer

(4)

\int_{0}^{a} \frac{x d x}{cos ( x ) cos ( a - x )} = - \frac{a}{sin ( a )} ln (cos (a))

Note that

(5)

d (ln (\frac{cos ( a - x )}{cos ( x )})) = \frac{sin ( a )}{cos ( x ) cos ( a - x )} d x

and

(6)

\int_{0}^{a} \frac{1}{sin ( a )} ln (\frac{cos ( a - x )}{cos ( x )}) d x = 0

Since the function is odd around $a / 2$ .

(7)

\int_{0}^{a} \frac{x d x}{cos ( x ) cos ( a - x )} = = = \int_{0}^{a} \frac{x}{sin ( a )} d (ln (\frac{cos ( a - x )}{cos ( x )})) [\frac{x}{sin ( a )} ln (\frac{cos ( a - x )}{cos ( x )})]_{0}^{a} - \int_{0}^{a} \frac{1}{sin ( a )} ln (\frac{cos ( a - x )}{cos ( x )}) d x - \frac{a}{sin ( a )} ln (cos (a))

Answer

(8)

\int exp (arccos x) d x = \frac{1}{2} (x - 1 - x^{2}) exp (arccos x) + C

By substituting $x = cos θ$ and noting:

(9)

\frac{d}{d θ} exp θ \cdot (sin θ - cos θ) = 2 exp θ \cdot sin θ

Answer

(10)

\int_{0}^{t} \frac{x}{1 - x ^{3}} d x = \frac{2}{3} arcsin \frac{3}{2} t

Classic substitution $x^{3} = u^{2}$ , so $x^{\frac{1}{2}} d x = \frac{2}{3} d u$ .

⚠ Might have got fractions the wrong way round ⚠

Answer

(11)

\int_{0}^{t} \frac{x ^{- \frac{1}{2}}}{1 + x ^{\frac{1}{3}}} d x = 6 (t^{\frac{1}{6}} - arctan t^{\frac{1}{6}})

Let $y^{6} = x$ , then $d x = 6 y^{5} d y$ . Let $T : = t^{\frac{1}{6}}$ .

(12)

\int_{0}^{t} \frac{x ^{- \frac{1}{2}}}{1 + x ^{\frac{1}{3}}} d x = 6 \int_{0}^{T} \frac{y ^{2}}{1 + y ^{2}} d y = 6 \int_{0}^{T} \frac{( y ^{2} + 1 ) - 1}{1 + y ^{2}} d y = 6 \int_{0}^{T} 1 - \frac{1}{1 + y ^{2}} d y = 6 [y - a t a n (y)]_{0}^{T}

Answer

(13)

\int \frac{x cosh x}{sinh ^{2} x} d x = lo g (tanh \frac{x}{2}) - \frac{x}{sinh x} + C