KindiakMath

Some Integration Drills

Problem 1. Evaluate the integral \displaystyle \int(x^2 + \tan x - 3)\, \mathrm dx.

(Click for Solution)

Solution. By the linearity of integration,

\displaystyle \begin{aligned}\int(x^2 + \tan x - 3)\, \mathrm dx &= \int x^2\, \mathrm dx + \int \tan x \, \mathrm dx - \int 3\, \mathrm dx \\ &= \frac 13 x^3 - \ln|{\cos x}| - 3x + C.\end{aligned}

Problem 2. Evaluate the integral \displaystyle \int \left(\frac{1}{9 + x^2} + \frac{1}{\sqrt{4-x^2}} \right)\, \mathrm dx.

(Click for Solution)

Solution. By the linearity of integration,

\displaystyle \begin{aligned} \int \left(\frac{1}{9 + x^2} + \frac{1}{\sqrt{4-x^2}} \right)\, \mathrm dx &= \int \frac{1}{9 + x^2} \, \mathrm dx + \int \frac{1}{ \sqrt{4-x^2} } \, \mathrm dx \\ &= \int \frac{1}{3^2 + x^2} \, \mathrm dx + \int \frac{1}{ \sqrt{2^2-x^2} } \, \mathrm dx \\ &= \frac 13 \tan^{-1} \left( \frac x3 \right) + \sin^{-1} \left( \frac x2 \right) + C. \end{aligned}

Problem 3. Evaluate the integral \displaystyle \int_0^1 x \sec(x^2) \tan(x^2)\, \mathrm dx.

(Click for Solution)

Solution. By the substitution \displaystyle u = x^2 \Rightarrow \mathrm dx = \frac 1{2x}\, \mathrm du,

\displaystyle \begin{aligned} \int x \sec(x^2) \tan(x^2)\, \mathrm dx &= \int x \sec(u) \tan(u) \cdot \frac{1}{2x}\, \mathrm du \\ &= \frac 12 \int \sec(u) \tan(u)\, \mathrm du \\ &= \frac 12 \sec(u) + C \\ &= \frac 12 \sec(x^2) + C. \end{aligned}

Hence, substituting the limits,

\displaystyle \begin{aligned} \int_0^1 x \sec(x^2) \tan(x^2)\, \mathrm dx &= \frac 12 [\sec(x^2)]_0^1\\ &= \frac 12 (\sec(1) - \sec(0)) \\ &= \frac 1{2 \cos(1)} - \frac 12 \\ &\approx 0.425, \end{aligned}

when approximated to 3 decimal places.

Problem 4. Evaluate the integral \displaystyle \int \tan^3 x\, \mathrm dx.

(Click for Solution)

Solution. Using the Pythagorean identity \sec^2 x = \tan^2 x +1,

\displaystyle \begin{aligned} \int \tan^3 x\, \mathrm dx &= \int \tan x \tan^2 x\, \mathrm dx \\ &= \int \tan x (\sec^2 x - 1)\, \mathrm dx \\ &= \int \tan x \sec^2 x \, \mathrm dx - \int \tan x\, \mathrm dx \\ &= \int \tan x \sec^2 x \, \mathrm dx -(- \ln|{\cos x}|) + C.\end{aligned}

For the first integral, make the substitution \displaystyle u = \tan x \Rightarrow \mathrm dx = \frac{1}{\sec^2 x}\, \mathrm du. Then

\displaystyle \begin{aligned}\int \tan x \sec^2 x \, \mathrm dx &= \int u \sec^2 x \cdot \frac{1}{\sec^2 x}\, \mathrm du \\ &= \int u\, \mathrm du \\ &= \frac 12 u^2 + C\\ &= \frac 12 \tan^2 x + C.\end{aligned}

Therefore, \displaystyle \int \tan^3 x\, \mathrm dx = \frac 12 \tan^2 x + \ln|{\cos x}| + C.

Problem 5. Evaluate the integral \displaystyle \int \sin(\ln \theta)\, \mathrm d\theta.

(Click for Solution)

Solution. Make the substitution u = \ln \theta \Rightarrow \mathrm d\theta = e^u\, \mathrm du. Then

\displaystyle \int \sin(\ln \theta)\, \mathrm d\theta = \int e^u \sin(u)\, \mathrm du.

Integrating by parts twice,

\displaystyle \begin{aligned}\int e^u \sin(u)\, \mathrm du &= \underbrace{e^u}_{\mathrm I}\, \underbrace{\sin(u)}_{\mathrm S} - \int \underbrace{e^u}_{\mathrm I}\, \underbrace{\cos(u)}_{\mathrm D}\, \mathrm du \\ &= e^u \sin(u) - \int e^u \cos(u)\, \mathrm du\\ &= e^u \sin(u) - \left(\underbrace{e^u}_{\mathrm I}\, \underbrace{\cos(u)}_{\mathrm S} - \int \underbrace{e^u}_{\mathrm I}\, \underbrace{-\sin(u)}_{\mathrm D}\, \mathrm du\right) \\ &= e^u \sin(u) - \left(e^u \cos(u) + \int e^u \sin(u) \, \mathrm du\right) \\ &= e^u \sin(u) - e^u \cos(u) - \int e^u \sin(u)\, \mathrm du \\&= e^u (\sin(u) - \cos(u)) - \int e^u \sin(u)\, \mathrm du.\end{aligned}

By algebruh,

\displaystyle \begin{aligned} \int e^u \sin(u)\, \mathrm du + \int e^u \sin(u)\, \mathrm du &= e^u (\sin(u) - \cos(u)) \\ 2 \int e^u \sin(u)\, \mathrm du &= e^u (\sin(u) - \cos(u)) \\ \int e^u \sin(u)\, \mathrm du &= \frac 12 e^u (\sin(u) - \cos(u)) + C\\ \int \sin(\ln \theta)\, \mathrm d\theta &= \frac 12 \theta (\sin(\ln \theta) - \cos(\ln \theta)) + C.\end{aligned}

Problem 6. Evaluate the integral \displaystyle \int \left(\frac{1}{36 - x^2} + \frac{1}{\sqrt{36-x^2}} \right)\, \mathrm dx.

(Click for Solution)

Solution. By the linearity of integration,

\displaystyle \begin{aligned} \int \left(\frac{1}{36 - x^2} + \frac{1}{\sqrt{36 -x^2}} \right)\, \mathrm dx &= \int \frac{1}{36 - x^2} \, \mathrm dx + \int \frac{1}{ \sqrt{36-x^2} } \, \mathrm dx \\ &= \int \frac{1}{6^2 - x^2} \, \mathrm dx + \int \frac{1}{ \sqrt{6^2-x^2} } \, \mathrm dx \\ &= \frac 1{2 \cdot 6} \ln \left| \frac{6+x}{6-x} \right| + \sin^{-1} \left( \frac x6 \right) + C\\ &= \frac 1{12} \ln \left| \frac{6+x}{6-x} \right| + \sin^{-1} \left( \frac x6 \right) + C. \end{aligned}

Problem 7. Evaluate the integral \displaystyle \int_0^2 x \sin(x^2 + 5)\, \mathrm dx.

(Click for Solution)

Solution. Making the substitution u = x^2 + 5 yields

\displaystyle \mathrm du = 2x\, \mathrm dx,\quad u(0) = 5, \quad u(2) = 9.

Therefore,

\displaystyle \begin{aligned} \int_0^2 x \sin(x^2 + 5)\, \mathrm dx &= \frac 12 \int_0^2 \sin(x^2 + 5) \cdot 2x\, \mathrm dx \\ &= \frac 1{2}\int_5^9 \sin(u)\, \mathrm du \\ &= \frac 12 [-\cos(u)]_5^9 \\ &= \frac 12 (-\cos(9) - (-\cos(5))) \\ &= \frac 12 (\cos(5) - \cos(9)) \approx 0.597, \end{aligned}

when approximated to 3 decimal places.

Problem 8. Evaluate the integral \displaystyle \int \cot^3 x\, \mathrm dx.

(Click for Solution)

Solution. Recall that \cot(x) = \tan(\pi/2 - x). Making the substitution u = \pi/2-x \Rightarrow \mathrm du = -\mathrm dx then using the result in Problem 4,

\displaystyle \begin{aligned} \int \cot^3 x\, \mathrm dx &= \int \tan^3 (\pi/2-x)\, \mathrm dx \\ &= \int \tan^3(u) \cdot -1\, \mathrm du \\ &= -\int \tan^3(u)\, \mathrm du \\ &= -\left( \frac 12 \tan^2 u + \ln|{\cos u}| \right) + C \\ &= -\left( \frac 12 \tan^2 (\pi/2-x) + \ln|{\cos (\pi/2-x)}| \right) + C \\ &= -\left( \frac 12 \cot^2 x + \ln|{\sin x}| \right) + C \\ &= - \frac 12 \cot^2 x - \ln|{\sin x}| + C \end{aligned}

Problem 9. Evaluate the integral \displaystyle \int e^{2\theta} \sin (e^\theta) \, \mathrm d\theta .

(Click for Solution)

Solution. Making the substitution \displaystyle u = e^{\theta} \Rightarrow \mathrm d\theta = \frac 1{u}\, \mathrm du then integrating by parts,

\displaystyle \begin{aligned} \int e^{2\theta} \sin(e^\theta) \, \mathrm d\theta &= \int u^2 \sin(u)\cdot \frac 1{u}\, \mathrm du \\ &= \int u \sin(u)\, \mathrm du\\ &= \underbrace{-\cos(u)}_{\mathrm I} \underbrace{u}_{\mathrm S} - \int \underbrace{-\cos(u)}_{\mathrm I}\underbrace{1}_{\mathrm D}\, \mathrm du \\ &= -u \cos u + \int \cos u\, \mathrm du\\ &= -u \cos u + \sin u + C\\ &= -e^\theta \cos(e^\theta) + \sin(e^\theta) + C. \end{aligned}

Problem 10. Given that

f(x) = \begin{cases} 1 - |x|, & |x| \leq 1, \\ 0, &\text{otherwise},\end{cases}

evaluate \displaystyle \int_0^{2} f(x^2-1)\, \mathrm dx.

(Click for Solution)

Solution. By the definition of f,

f(x^2 - 1) = \begin{cases} 1 - |x^2 - 1|, & |x^2 - 1| \leq 1, \\ 0, &\text{otherwise}.\end{cases}

Parsing the inequality |x^2 - 1| \leq 1 \iff |x| \leq \sqrt 2,

f(x^2 - 1) = \begin{cases} 1 - |x^2 - 1|, & |x| \leq \sqrt 2, \\ 0, &\text{otherwise}.\end{cases}

By definition of the absolute value,

|x^2 - 1| = \begin{cases} x^2 - 1, & |x| \geq 1, \\ 1 - x^2, &|x| \leq 1.\end{cases}

Parsing the inequalities again,

f(x^2 - 1) = \begin{cases} x^2, & 0 \leq |x| \leq 1, \\ 2-x^2 , & 1 \leq |x| \leq \sqrt 2, \\ 0, &\text{otherwise}.\end{cases}

Therefore, we can evaluate the integral as

\begin{aligned} \displaystyle \int_0^{2} f(x^2-1)\, \mathrm dx &= \int_0^1 f(x^2 - 1)\, \mathrm dx + \int_1^{\sqrt 2} f(x^2 - 1)\, \mathrm dx + \int_{\sqrt 2}^{2} 0\, \mathrm dx \\ &= \int_0^1 x^2\, \mathrm dx + \int_1^{\sqrt 2} (2 - x^2)\, \mathrm dx+ 0 \\ &= \left[ \frac{x^3}{3}\right]_0^1 + \left[ 2x - \frac{x^3}{3} \right]_1^{\sqrt 2} \\ &= \left( \frac 13 - 0 \right) + \left( 2\sqrt 2 - \frac{2\sqrt 2}{3} \right) \\ &= \frac 13 + \frac{4}{3}\sqrt 2. \end{aligned}

—Joel Kindiak, 20 Jan 25, 0925H

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