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Section 5.5 Indefinite Integrals and the Substitution Method

Example 5.5.1.

Let \(y=\sin(x^2+3)\text{,}\) so \(\frac{dy}{dx} = 2x \cos(x^2+3)\text{.}\)

To find \(\int 2x \cos(x^2+3)\, dx\text{,}\) it's useful to substitute \(u=x^2+3\) and \(du=2x\, dx\text{.}\)

So that \(\int 2x \cos(x^2+3)\, dx = \int \cos u \, du = \sin u + C = \sin(x^2+3) + C\text{.}\)

Example 5.5.3.

  1. \(\displaystyle \displaystyle\int 8(8x+32)^9\, dx\)

  2. \(\displaystyle \displaystyle\int \frac{x^3}{(8-x^4)^2}\, dx\)

  3. \(\displaystyle \displaystyle\int 8\csc^2 (4x) \cot (4x)\, dx\)

Answer.

  1. \(\displaystyle \frac{1}{10}(8x+32)^{10}+C\)

  2. \(\displaystyle \displaystyle\frac{1}{4(8-x^4)} + C\)

  3. \(\displaystyle -\cot^2 (4x) + C\)

Example 5.5.4.

  1. \(\displaystyle \displaystyle\int 2y^6 \sqrt{3-2y^7}\, dy\)

  2. \(\displaystyle \displaystyle\int\frac{3}{\sqrt{x}(7+3\sqrt{x})^2}\, dx\)

  3. !!\(\displaystyle\int \sin^8 (2x) \cos (2x)\, dx\)

  4. !!\(\displaystyle\int \sin^2 \theta \, d\theta\text{,}\) uses \(\sin^2\theta = \frac{1-\cos(2\theta)}{2}\text{.}\)

  5. !!\(\displaystyle\int x^3 \sqrt{x^2+7}\, dx\text{;}\) with \(u=x^2+7\text{,}\) \(x^2=u-7\text{,}\) so it becomes \(\int (u-7)\sqrt{u}\, du\)

  6. \(\displaystyle \displaystyle\int \frac{4}{(4x+3)\ln(4x+3)}\, dx\)

  7. !!\(\displaystyle\int\frac{21}{25+4r^2}\, dr\)

  8. !!\(\displaystyle\int \frac{18\tan^2 x \sec^2 x}{(2+\tan^3 x)^2}\, dx\text{;}\) substitute \(u=\tan x\text{,}\) then \(v=2+u^3\text{.}\)