[MATPY00] Integration by Substitution

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Integrate the following functions.

1. \(y(x) = x^2 + 5x + 9\).

SOLUTION. The integral \(Y(x)\) can be written as three separate indefinite integrations, 

\(Y(x) = \int x^2 dx + 5 \int x dx + 9 \int dx = \boxed{\frac{1}{3}x^3 + \frac{5}{2}x^2 + 9x + C }\textrm{   } \blacksquare\).

2. \(R(\theta) = \sin^2 \theta \cos \theta\).

SOLUTION. Let \(u = \sin \theta\) such that \(du = \cos \theta \textrm{ } d\theta\). Thus, 

\(\int \sin^2\theta \cos \theta \textrm{ } d\theta = \int u^2 du = \frac{1}{3}u^3 + C = \boxed{\frac{1}{3}\sin^3 \theta + C} \textrm{   } \blacksquare\).

3. \(Z(x) = 5x^2 \sec^2\left(x^3\right) \tan\left(x^3\right) \).

SOLUTION. Let \(u = \tan\left(x^3\right)\), such that \(du = 3x^2 \sec^2 \left(x^3\right) dx\). 

\(\int 5x^2 \sec^2\left(x^3\right) \tan\left(x^3\right) dx = \frac{5}{3} \int u du = \boxed{\frac{5}{6}\tan^2 \left(x^3\right) + C} \textrm{   } \blacksquare\). 

4. \(A(y) = 2y \log_{10} y^2\). 

SOLUTION. Let \(u = y^2\), so \(du = 2ydy\). 

\(\int 2y \log_{10} y^2 dy = \int \log_{10} u du = \frac{1}{\ln 10}\int \ln u du = \boxed{\frac{1}{\ln 10} y^2 \ln y^2 - \frac{1}{\ln 10} y^2} \textrm{   } \blacksquare\). 

5. \(f(x) = \frac{10 \sqrt{x}}{x} \exp\left(\sqrt{x}\right)\). 

SOLUTION. Let \(u = \sqrt{x}\), and \(du = \frac{1}{2\sqrt{x}}dx\). 

\(\int \frac{10 \sqrt{x}}{x} \exp\left(\sqrt{x}\right) = 20 \int e^u du = \boxed{20 e^{\sqrt{x}}} \textrm{   } \blacksquare\).