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Notes

Practice Problems

Find the indefinite integral

\(\textbf{1)}\)\(\displaystyle\int \sin ⁡ x \,dx \)

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The answer is \(-\cos ⁡x+C\)

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\(\,\,\,\,\,\displaystyle\int \sin ⁡ x \,dx=-\cos ⁡x+C \,\,\,\left(\text{from the notes}\right)\)

\(\textbf{2)}\)\(\displaystyle\int \cos ⁡ x \,dx \)

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The answer is \(\sin ⁡x+C\)

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\(\,\,\,\,\,\displaystyle\int \cos ⁡ x \,dx=\sin ⁡x+C \,\,\,\left(\text{from the notes}\right)\)

\(\textbf{3)}\)\(\displaystyle\int(5 \sin ⁡x-2 \cos ⁡x) \,dx \)

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The answer is \(-5 \cos ⁡x-2 \sin ⁡x+C\)

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\(\,\,\,\,\,\displaystyle\int(5 \sin ⁡x-2 \cos ⁡x) \,dx\)

\(\,\,\,\,\,\displaystyle\int 5 \sin ⁡x \,dx – \int 2 \cos ⁡x \,dx\)

\(\,\,\,\,\,\displaystyle 5 \cdot \int \sin ⁡x \,dx – 2 \cdot \int \cos ⁡x \,dx\)

\(\,\,\,\,\,\displaystyle 5 \left(-\cos x +C_1\right) – 2 \left(\sin x + C_2\right)\)

\(\,\,\,\,\,\displaystyle -5\cos x +5C_1 – 2\sin x – 2C_2\)

\(\,\,\,\,\,\displaystyle -5\cos x – 2\sin x + 5C_1 – 2C_2\)

\(\,\,\,\,\,\displaystyle -5\cos x – 2\sin x + C\)

\(\textbf{4)}\)\(\displaystyle\int\left(\sec^2 ⁡ x – \csc^2 ⁡x \right) \,dx \)

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The answer is \(\tan ⁡x+\cot ⁡x+C\)

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\(\,\,\,\,\,\displaystyle\int\left(\sec^2 ⁡ x – \csc^2 ⁡x \right) \,dx \)

\(\,\,\,\,\,\displaystyle\int\sec^2 ⁡ x – \int \csc^2 ⁡x \,dx \)

\(\,\,\,\,\,\tan ⁡x + C_1 ⁡- \left(-\cot ⁡x + C_2\right) \)

\(\,\,\,\,\,\tan ⁡x + C_1 +\cot ⁡x – C_2 \)

\(\,\,\,\,\,\tan ⁡x +\cot ⁡x + C_1 – C_2 \)

\(\,\,\,\,\,\tan ⁡x +\cot ⁡x + C \)

\(\textbf{5)}\)\(\displaystyle\int5 \sec ⁡x \tan ⁡x \,dx \)

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The answer is \(5 \sec ⁡x+C\)

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\(\,\,\,\,\,\displaystyle\int5 \sec ⁡x \tan ⁡x \,dx \)

\(\,\,\,\,\,\displaystyle 5 \cdot \int\sec ⁡x \tan ⁡x \,dx \)

\(\,\,\,\,\,5 \sec ⁡x+C \)

\(\textbf{6)}\)\(\displaystyle\int -4\sin ⁡ x \,dx \)

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The answer is \(4\cos ⁡x+C\)

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\(\,\,\,\,\,\displaystyle\int -4\sin ⁡ x \,dx \)

\(\,\,\,\,\,\displaystyle-4 \cdot \int \sin ⁡ x \,dx \)

\(\,\,\,\,\,-4 \left(-\cos x + C_1\right) \)

\(\,\,\,\,\,4\cos x + 4C_1 \)

\(\,\,\,\,\,4\cos x + C \)

\(\textbf{7)}\)\(\displaystyle\int 8 \sec^2 ⁡ x \,dx \)

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The answer is \(8\tan ⁡x+C\)

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\(\,\,\,\,\,\displaystyle\int 8 \sec^2 ⁡ x \,dx \)

\(\,\,\,\,\,\displaystyle 8 \cdot \int \sec^2 ⁡ x \,dx \)

\(\,\,\,\,\,8 \left(\tan ⁡x+C_1\right) \)

\(\,\,\,\,\,8\tan ⁡x+ 8 C_1 \)

\(\,\,\,\,\,8\tan ⁡x+ C \)

\(\textbf{8)}\)\(\displaystyle\int -5\csc ⁡ x \cot ⁡ x \,dx \)

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The answer is \(5\csc ⁡x+C\)

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\(\,\,\,\,\,\displaystyle\int -5\csc ⁡ x \cot ⁡ x \,dx \)

\(\,\,\,\,\,\displaystyle -5 \cdot \int \csc ⁡ x \cot ⁡ x \,dx \)

\(\,\,\,\,\,\displaystyle -5\left(-\csc x +C_1\right) \)

\(\,\,\,\,\,\displaystyle 5\csc x -5C_1 \)

\(\,\,\,\,\,\displaystyle 5\csc x +C \)

\(\textbf{9)}\)\(\displaystyle\int \frac{5}{ \sec ⁡ x} \,dx \)

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The answer is \(5\sin ⁡x+C\)

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\(\,\,\,\,\,\displaystyle\int \frac{5}{ \sec ⁡ x} \,dx \)

\(\,\,\,\,\,\displaystyle\int 5 \cos x \,dx \)

\(\,\,\,\,\,\displaystyle 5 \cdot \int \cos x \,dx \)

\(\,\,\,\,\, 5 \left(\sin x +c_1 \right) \)

\(\,\,\,\,\, 5\sin x +5c_1 \)

\(\,\,\,\,\, 5\sin x + C \)

\(\textbf{10)}\)\(\displaystyle\int \frac{15}{ \csc ⁡ x} \,dx \)

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The answer is \(-15\cos ⁡x+C\)

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\(\,\,\,\,\,\displaystyle\int \frac{15}{ \csc ⁡ x} \,dx \)

\(\,\,\,\,\,\displaystyle\int 15\sin⁡ x \,dx \)

\(\,\,\,\,\,\displaystyle 15 \cdot \int \sin⁡ x \,dx \)

\(\,\,\,\,\, 15 \left(-\cos x + C_1\right) \)

\(\,\,\,\,\, -15\cos x + 15C_1 \)

\(\,\,\,\,\, -15\cos x + C \)

\(\textbf{11)}\)\(\displaystyle\int 10 \tan ⁡ x \,dx \)

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The answer is \(-10 \ln |\cos ⁡x|+C\)

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\(\,\,\,\,\,\displaystyle\int 10 \cdot \tan ⁡ x \,dx \)

\(\,\,\,\,\,\displaystyle 10 \cdot \int \tan ⁡ x \,dx \)

\(\,\,\,\,\,\displaystyle 10 \cdot \int \frac{\sin x}{\cos x} \,dx \)

\(\,\,\,\,\,u=\cos x, \,\,\, du=-\sin x \,dx \)

\(\,\,\,\,\,\displaystyle 10 \cdot \int -\frac{1}{u} \,du \)

\(\,\,\,\,\,\displaystyle -10 \cdot \int \frac{1}{u} \,du \)

\(\,\,\,\,\,\displaystyle \int \frac{1}{u} \,du=\ln \left(\left|u\right|\right) \left(\text{notes}\right) \)

\(\,\,\,\,\, -10\left(\ln \left|u\right|\right) + C \)

\(\,\,\,\,\, -10 \ln \left|\cos x \right|+ C \)

\(\textbf{12)}\)\(\displaystyle\int \cot ⁡ x \,dx \)

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The answer is \(\ln |\sin ⁡x|+C\)

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\(\,\,\,\,\,\displaystyle\int \cot ⁡ x \,dx \)

\(\,\,\,\,\,\displaystyle\int \frac{\cos \left(x\right)}{\sin \left(x\right)}\,dx \)

\(\,\,\,\,\,\displaystyle\int \frac{\cos \left(x\right)}{\sin \left(x\right)}\,dx \)

\(\,\,\,\,\,u=\sin x, \,\,\, du=\cos x \,dx \)

\(\,\,\,\,\,\displaystyle\int \frac{1}{u} \,du \)

\(\,\,\,\,\,\displaystyle \int \frac{1}{u} \,du=\ln \left(\left|u\right|\right) \left(\text{notes}\right) \)

\(\,\,\,\,\,\ln \left|u\right| + C \)

\(\,\,\,\,\,\ln \left|\sin x\right| + C \)

Challenge Problems

\(\textbf{13)}\)\(\displaystyle\int 5\csc ⁡ x \,dx \)

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The answer is \(-5\ln |\csc ⁡x + \cot x| +C\)

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\(\,\,\,\,\,\displaystyle\int 5\csc ⁡ x \,dx \)

\(\,\,\,\,\,\displaystyle5 \cdot \int \csc ⁡ x \,dx \)

\(\,\,\,\,\,\displaystyle5 \cdot \int \frac{\left(\csc x \right)\left(\csc x + \cot x\right)}{\left(\csc x + \cot x\right)} \,dx \)

\(\,\,\,\,\,\displaystyle5 \cdot \int \frac{\csc^2 x + \cot x \csc x}{\left(\csc x + \cot x\right)} \,dx \)

\(\,\,\,\,\,u=\csc x + \cot x \,\,\, du= -\csc^2x -\cot x \csc x \,dx\)

\(\,\,\,\,\,\displaystyle -\int \frac{1}{u} \,du \)

\(\,\,\,\,\,\displaystyle \int \frac{1}{u} \,du=\ln \left(\left|u\right|\right) \left(\text{notes}\right) \)

\(\,\,\,\,\,-\ln \left(\left|u\right|\right) + C \)

\(\,\,\,\,\,-5\ln \left|\csc x +\cot x\right| + C \)

\(\textbf{14)}\)\(\displaystyle\int \cos ⁡ x \sin x \,dx \)

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The answer is \(\displaystyle\frac{\sin^2 x}{2}+C\)

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\(\,\,\,\,\,\displaystyle\int \cos ⁡ x \sin x \,dx \)

\(\,\,\,\,\,u=\sin x, \,\,\, du=\cos x \,dx\)

\(\,\,\,\,\,\displaystyle\int u \,du \)

\(\,\,\,\,\,\displaystyle \frac{u^2}{2} +C \)

\(\,\,\,\,\,\displaystyle \frac{\sin^2 x}{2} +C \)

\(\textbf{15)}\)\(\displaystyle\int x \sin x \,dx \)

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The answer is \(-x\cos ⁡x +\sin x+C\)

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\(\,\,\,\,\,\displaystyle\int x \sin x \,dx \)

\(\,\,\,\,\,\displaystyle\int fg’=fg-\int f’g \,\,\, \left(\text{Integration by parts}\right)\)

\(\,\,\,\,\,f=x, \,\,\, g’=\sin x\)

\(\,\,\,\,\,f’=1, \,\,\, g=-\cos x\)

\(\,\,\,\,\,\displaystyle\int x \sin x \,dx = -x \cos x – \int -\cos x \, dx \)

\(\,\,\,\,\,\displaystyle\int x \sin x \,dx = -x \cos x + \int \cos x \, dx \)

\(\,\,\,\,\,\displaystyle\int x \sin x \,dx = -x \cos x + \sin x + C \)

\(\textbf{16)}\)\(\displaystyle\int -4 \cot x \csc^4 x \,dx \)

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The answer is \(\csc^4 ⁡x+C\)

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\(\,\,\,\,\,\displaystyle\int -4 \cot x \csc^4 x \,dx \)

\(\,\,\,\,\,u=\csc^4x\)

\(\,\,\,\,\,du= 4 \csc^3 x \cdot \left(-\csc x \cot x \right)= -4 \csc^4 x \cot x\)

\(\,\,\,\,\,\displaystyle\int du \)

\(\,\,\,\,\,u + C \)

\(\,\,\,\,\, \csc^4x + C \)

See Related Pages\(\)

\(\bullet\text{ Calculus Homepage}\)
\(\,\,\,\,\,\,\,\,\text{All the Best Topics…}\)

\(\bullet \text{ Indefinite Integral Calculator (Symbolab)}\)

\(\bullet\text{ Trapezoidal Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{b-a}{2n}\left[f(a)+2f(x_1)+2f(x_2)+…+2fx_{n-1}+f(b)\right]…\)

\(\bullet\text{ Properties of Integrals}\)
\(\,\,\,\,\,\,\,\,\displaystyle \int_{a}^{b}cf(x) \, dx=c\displaystyle \int_{a}^{b}f(x) \,dx…\)

\(\bullet\text{ Indefinite Integrals- Power Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle \int x^n \, dx = \displaystyle \frac{x^{n+1}}{n+1}+C…\)

\(\bullet\text{ Indefinite Integrals- Trig Functions}\)
\(\,\,\,\,\,\,\,\,\displaystyle \int \cos{x} \, dx=\sin{x}+C…\)

\(\bullet\text{ Definite Integrals}\)
\(\,\,\,\,\,\,\,\,\displaystyle \int_{5}^{7} x^3 \, dx…\)

\(\bullet\text{ Integration by Substitution}\)
\(\,\,\,\,\,\,\,\,\displaystyle \int (x^2+3)^3(2x) \,dx…\)

\(\bullet\text{ Area of Region Between Two Curves}\)
\(\,\,\,\,\,\,\,\,A=\displaystyle \int_{a}^{b}\left[f(x)-g(x)\right]\,dx…\)

\(\bullet\text{ Arc Length}\)
\(\,\,\,\,\,\,\,\,\displaystyle \int_{a}^{b}\sqrt{1+\left[f'(x)\right]^2} \,dx…\)

\(\bullet\text{ Average Function Value}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{1}{b-a} \int_{a}^{b}f(x) \,dx\)

\(\bullet\text{ Volume by Cross Sections}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Disk Method}\)
\(\,\,\,\,\,\,\,\,V=\displaystyle \int_{a}^{b}\left[f(x)\right]^2\,dx…\)

\(\bullet\text{ Cylindrical Shells}\)
\(\,\,\,\,\,\,\,\,V=2 \pi \displaystyle \int_{a}^{b} y f(y) \, dy…\)

In Summary

Indefinite integrals, also known as antiderivatives, are a fundamental concept in calculus that allow us to find the original function when given its derivative. The derivatives and antiderivatives of trig functions are in terms of other trig functions. Memorizing or having the notes for the basic trig derivatives can help a lot in evaluating these types of integrals.

Indefinite integrals are covered in calculus courses after limits and derivatives are introduced. The trigonometric indefinite integrals are usually memorized by students, but they can be derived in the course material too. Some related topics to indefinite integrals include definite integrals, integration by substitution, integration by parts, and the fundamental theorem of calculus. Understanding these concepts can help students get a deeper understanding of indefinite integrals and how they can be used to solve real-world problems.

The trigonometric functions sine, cosine, tangent, secant, cosecant and cotangent are initially introduced to study angles and sides in triangles. But eventually we notice trig functions show up in a lot of math, specifically in studying periodic phenomena.

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