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Notes

Practice Problems

Find the integral

\(\textbf{1)}\)\(\displaystyle \int (x^2+3)^3(2x) \,dx\)

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\(\text{The answer is } \frac{1}{4} (x^2+3)^4+C\)

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\(\,\,\,\,\,u=x^2+3\)

\(\,\,\,\,\,du=2x \,dx\)

\(\,\,\,\,\,\displaystyle \int u^3 \,du\)

\(\,\,\,\,\,\displaystyle \frac{u^4}{4}+C\)

\(\,\,\,\,\,\)The answer is \(\frac{1}{4} (x^2+3)^4+C\)

\(\textbf{2)}\)\(\displaystyle \int x^2(5x^3+3)^2 \,dx\)

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\(\text{The answer is } \frac{25}{9}x^9+5x^6+3x^3+C\)

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\(\,\,\,\,\,u=5x^3+3\)

\(\,\,\,\,\,du=15x^2 \,dx\)

\(\,\,\,\,\,\frac{1}{15}du=x^2\, dx\)

\(\,\,\,\,\,\displaystyle \frac{1}{15}\int u^2 \,du\)

\(\,\,\,\,\,\displaystyle \frac{1}{15} \frac{u^3}{3}+C\)

\(\,\,\,\,\,\displaystyle \frac{1}{15} \frac{\left(5x^3+3\right)^3}{3}+C\)

\(\,\,\,\,\,\)The answer is \(\frac{25}{9}x^9+5x^6+3x^3+C\)

\(\textbf{3)}\)\(\displaystyle \int \frac{\sin x}{\cos^4 x} \,dx\)

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\(\text{The answer is } \displaystyle \frac{1}{3 \cos^3 ⁡x }+C\)

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\(\,\,\,\,\,u=\cos x\)

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

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

\(\,\,\,\,\,\displaystyle -\int \frac{1}{u^4} \,du\)

\(\,\,\,\,\,\displaystyle -\int u^{-4} \,du\)

\(\,\,\,\,\,\displaystyle -\frac{u^{-3}}{-3}+C\)

\(\,\,\,\,\,\displaystyle \frac{u^{-3}}{3}+C\)

\(\,\,\,\,\,\displaystyle \frac{\left(\cos x\right)^{-3}}{3}+C\)

\(\,\,\,\,\,\)The answer is \(\displaystyle \frac{1}{3 \cos^3 ⁡x }+C\)

\(\textbf{4)}\)\(\displaystyle \int x \sin (x^2) \,dx\)

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\(\text{The answer is } \displaystyle -\frac{1}{2} \cos⁡(x^2)+C\)

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\(\,\,\,\,\,u=x^2\)

\(\,\,\,\,\,du=2x \,dx\)

\(\,\,\,\,\,\frac{1}{2}du= x \, dx\)

\(\,\,\,\,\,\displaystyle \frac{1}{2} \int \sin u \,du\)

\(\,\,\,\,\,\displaystyle \frac{1}{2} (-\cos u) +C\)

\(\,\,\,\,\,\)The answer is \(-\frac{1}{2} \cos⁡(x^2)+C\)

\(\textbf{5)}\)\(\displaystyle \int \frac{8x^4}{\sqrt{x^5-2}} \,dx\)

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\(\text{The answer is } \displaystyle \frac{16}{5} \sqrt{x^5-2}+C\)

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\(\,\,\,\,\,u=x^5-2\)

\(\,\,\,\,\,du=5x^4 \,dx\)

\(\,\,\,\,\,\frac{1}{5}du= x^4 \, dx\)

\(\,\,\,\,\,\displaystyle \frac{1}{5} \int \frac{8}{\sqrt{u}} \,du\)

\(\,\,\,\,\,\displaystyle \frac{8}{5} \int u^{-1/2} \,du\)

\(\,\,\,\,\,\displaystyle \frac{8}{5} \cdot 2u^{1/2} +C\)

\(\,\,\,\,\,\displaystyle \frac{16}{5} \sqrt{u} +C\)

\(\,\,\,\,\,\)The answer is \(\frac{16}{5} \sqrt{x^5-2}+C\)

\(\textbf{6)}\)\(\displaystyle \int 5x^2 \sqrt{x^3+10} \,dx\)

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\(\text{The answer is } \displaystyle \frac{10}{9} (x^3+10)^{3/2}+C\)

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\(\,\,\,\,\,u=x^3+10\)

\(\,\,\,\,\,du=3x^2 \,dx\)

\(\,\,\,\,\,\frac{1}{3}du= x^2 \, dx\)

\(\,\,\,\,\,\displaystyle \frac{1}{3} \int 5 \sqrt{u} \,du\)

\(\,\,\,\,\,\displaystyle \frac{5}{3} \int u^{1/2} \,du\)

\(\,\,\,\,\,\displaystyle \frac{5}{3} \cdot \frac{2}{3} u^{3/2} +C\)

\(\,\,\,\,\,\displaystyle \frac{10}{9} u^{3/2} +C\)

\(\,\,\,\,\,\)The answer is \(\frac{10}{9} (x^3+10)^{3/2}+C\)

\(\textbf{7)}\)\(\displaystyle \int \frac{\sec^2 (\frac{1}{x^6})}{x^7} \,dx\)

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\(\text{The answer is } \displaystyle -\frac{1}{6} \tan⁡ (\frac{1}{x^6})+C\)

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\(\,\,\,\,\,u=\displaystyle \frac{1}{x^6}\)

\(\,\,\,\,\,u=\displaystyle x^{-6}\)

\(\,\,\,\,\,du=-6x^{-7} \,dx\)

\(\,\,\,\,\,\displaystyle -\frac{1}{6}du=x^{-7} \, dx\)

\(\,\,\,\,\,\displaystyle -\frac{1}{6}du=\frac{1}{x^{7}} \, dx\)

\(\,\,\,\,\,\displaystyle -\frac{1}{6} \int \sec^2 u \,du\)

\(\,\,\,\,\,\displaystyle -\frac{1}{6} \tan u +C\)

\(\,\,\,\,\,\)The answer is \(-\frac{1}{6} \tan⁡ (\frac{1}{x^6})+C\)

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

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\(\text{The answer is } \displaystyle \frac{1}{3} \tan^3⁡ (x) +C\)

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\(\,\,\,\,\,u=\tan x\)

\(\,\,\,\,\,du= \sec^2 x \,dx\)

\(\,\,\,\,\,\displaystyle \int u^2 \,du\)

\(\,\,\,\,\,\displaystyle \frac{u^3}{3} +C\)

\(\,\,\,\,\,\)The answer is \(\frac{1}{3} \tan^3⁡ (x) +C\)

\(\textbf{9)}\)\(\displaystyle \int e^x \sqrt{15+e^x} \,dx\)

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\(\text{The answer is } \displaystyle \frac{2}{3} (15+e^x )^{3/2} +C\)

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\(\,\,\,\,\,u=15+e^x\)

\(\,\,\,\,\,du= e^x \,dx\)

\(\,\,\,\,\,\displaystyle \int \sqrt{u} \,du\)

\(\,\,\,\,\,\displaystyle \int u^{1/2} \,du\)

\(\,\,\,\,\,\displaystyle \frac{2u^{3/2}}{3} +C\)

\(\,\,\,\,\,\)The answer is \(\frac{2}{3} (15+e^x )^{3/2} +C\)

\(\textbf{10)}\)\(\displaystyle \int e^{\cos 5 \theta} \sin 5\theta \,d\theta\)

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\(\text{The answer is } \displaystyle -\frac{1}{5} e^{\cos(5\theta)} +C\)

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\(\,\,\,\,\,u=\cos (5 \theta)\)

\(\,\,\,\,\,du= -5\sin (5\theta) \,dx\)

\(\,\,\,\,\,-\frac{1}{5}du= \sin (5\theta) \, dx\)

\(\,\,\,\,\,\displaystyle -\frac{1}{5} \int e^u \,du\)

\(\,\,\,\,\,\displaystyle -\frac{1}{5}e^u +C\)

\(\,\,\,\,\,\)The answer is \(-\frac{1}{5} e^{\cos(5\theta)} +C\)

\(\textbf{11)}\)\(\displaystyle \int \frac{\sin (\ln(4x))}{x} \,dx\)

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\(\text{The answer is } \displaystyle -\cos⁡(\ln⁡(4x))+C\)

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\(\,\,\,\,\,u=\ln(4x)\)

\(\,\,\,\,\,du=\displaystyle \frac{1}{x} \,dx\)

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

\(\,\,\,\,\,\displaystyle -\cos u +C\)

\(\,\,\,\,\,\)The answer is \(-\cos⁡(\ln⁡(4x))+C\)

\(\textbf{12)}\)\(\displaystyle \int x e^{x^2} \,dx\)

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\(\text{The answer is } \displaystyle \frac{1}{2} e^{x^2} +C\)

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\(\,\,\,\,\,u=x^2\)

\(\,\,\,\,\,du=2x \,dx\)

\(\,\,\,\,\,\frac{1}{2}du= x \, dx\)

\(\,\,\,\,\,\displaystyle \frac{1}{2} \int e^u \,du\)

\(\,\,\,\,\,\displaystyle \frac{1}{2} e^u +C\)

\(\,\,\,\,\,\)The answer is \(\frac{1}{2} e^{x^2} +C\)

\(\textbf{17)}\)\(\displaystyle \int \frac{x}{\sqrt{x^2-7}} \,dx\)

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\(\text{The answer is } \sqrt{x^2-7}+C\)

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\(\text{Substitution: Let }u = x^2 – 7 \text{. This simplifies the square root expression.}\)
\(\,\,\,\,\,u=x^2-7\)

\(\text{Differentiate }u \text{ to find }du \text{. We need to match this with the }x \text{ dx term in the original integral.}\)
\(\,\,\,\,\,du=2x \,dx\)

\(\text{Divide by }2 \text{ to match the }x dx \text{ in the original integral.}\)
\(\,\,\,\,\,\frac{1}{2}du=x \,dx\)

\(\text{Substitute into the integral: replace }x dx \text{ with }(1/2) du \text{, and substitute }u \text{ for }x^2 – 7 \text{ in the square root.}\)
\(\,\,\,\,\,\displaystyle \frac{1}{2} \int \frac{1}{\sqrt{u}} \,du\)

\(\text{The integral of }\frac{1}{\sqrt{u}} \text{ is }2\sqrt{u} \text{. Multiply by the constant } \text{(1/2).}\)
\(\,\,\,\,\,\displaystyle \frac{1}{2} \cdot 2\sqrt{u}+C\)

\(\text{Simplify the expression, resulting in }\sqrt{u} \text{.}\)
\(\,\,\,\,\,\displaystyle \sqrt{u}+C\)

\(\text{Finally, substitute }u = x^2 – 7 \text{ back into the equation to get the final answer.}\)
\(\,\,\,\,\,\displaystyle \sqrt{x^2-7}+C\)

\(\,\,\,\,\,\)The answer is \(\sqrt{x^2-7}+C\)

See Related Pages\(\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

In Summary

Integration by Substitution (aka “u substitution”) is a method for simplifying certain tricky integrals (or antiderivatives.) It involves substituting a new variable for a certain part of the function being integrated. Usually you want to set it up so that the entire integral can be expressed in terms of your new variable “u” and the derivative of your new variable, “du.” They are commonly used with Trig functions, exponential functions, logarithmic functions, polynomial functions, and more.

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