A) Find the indefinite integral. (Use C for the constant of integration.) Integral of sin^3...

Question:

A) Find the indefinite integral. (Use {eq}C {/eq} for the constant of integration.)

{eq}\int \sin^3 (5 \theta) \sqrt{\cos(5\theta)} \, \mathrm{d} \theta {/eq}

B) Find the indefinite integral. (Use {eq}C {/eq} for the constant of integration.)

{eq}\int \frac{\cos^5 t}{\sqrt{\sin t}} \, \mathrm{d}t {/eq}

Integration by Substitution:

The given integration can be solved by following substitution method

{eq}\displaystyle \int p(q(y)) q'(y) y = P(q(y))+C. {/eq}

Where C is the constant of the integration.

Let's take simple example of {eq}\displaystyle q(y) = \cos y, {/eq} then {eq}\displaystyle P(y) = \sin y {/eq} and {eq}\displaystyle q(y) = y. {/eq} Then it becomes

{eq}\displaystyle \int \cos y \cdot 1 dy = \sin y + C. {/eq}

Remember following rules of integration:

1. Take the constant out: {eq}\displaystyle \int v\cdot f\left(v\right)dv=n\cdot \int f\left(v\right)dv. {/eq}

2. An algebraic property: {eq}\displaystyle \sin ^3\left(u\right)=\sin ^2\left(u\right)\sin \left(u\right). {/eq}

3. Trigonometric identity: {eq}\displaystyle \sin ^2\left(x\right)=1-\cos ^2\left(x\right). {/eq}

4. The sum rule: {eq}\displaystyle \int f\left(v\right)\pm g\left(v\right)dv=\int f\left(v\right)dv\pm \int g\left(v\right)dv. {/eq}

5. Radical rule: {eq}\displaystyle \sqrt{a}=a^{\frac{1}{2}}. {/eq}

6. The power rule: {eq}\displaystyle \int v^ndv=\frac{v^{n+1}}{n+1}, \quad n\ne -1. {/eq}

Answer and Explanation:

A) We have to solve the integration of $$\displaystyle I = \int \sin^3 (5 \theta) \sqrt{\cos(5\theta)} \, \mathrm{d} \theta $$

Apply substitution for {eq}u = 5 \theta \Rightarrow du = 5 d\theta. {/eq}

$$\displaystyle = \int \sin ^3\left(u\right)\sqrt{\cos \left(u\right)}\frac{1}{5}du $$

Take the constant out.

$$\displaystyle = \frac{1}{5}\cdot \int \sin ^3\left(u\right)\sqrt{\cos \left(u\right)}du $$

Use the algebraic property.

$$\displaystyle = \frac{1}{5}\cdot \int \sin ^2\left(u\right)\sin \left(u\right)\sqrt{\cos \left(u\right)}du $$

Use the trigonometric identity.

$$\displaystyle = \frac{1}{5}\cdot \int \left(1-\cos ^2\left(u\right)\right)\sin \left(u\right)\sqrt{\cos \left(u\right)}du $$

Apply substitution for {eq}v=\cos \left(u\right) \Rightarrow dv=-\sin \left(u\right) du. {/eq}

$$\displaystyle = \frac{1}{5}\cdot \int -\sqrt{v}\left(1-v^2\right)dv $$

Expand:

$$\displaystyle = \frac{1}{5}\cdot \int -\sqrt{v}+v^{\frac{5}{2}}dv $$

Apply the sum rule.

$$\displaystyle = \frac{1}{5}\left(-\int \sqrt{v}dv+\int v^{\frac{5}{2}}dv\right) $$

Use the radical rule.

$$\displaystyle = \frac{1}{5}\left(-\int v^{\frac{1}{2}}dv+\int v^{\frac{5}{2}}dv\right) $$

Apply the power rule.

$$\displaystyle = \frac{1}{5}\left(-\frac{2}{3}v^{\frac{3}{2}}+\frac{2}{7}v^{\frac{7}{2}}\right)+C $$

Substitute back {eq}v=\cos \left(u\right). {/eq}

$$\displaystyle = \frac{1}{5}\left(-\frac{2}{3}\cos ^{\frac{3}{2}}\left(u\right)+\frac{2}{7}\cos ^{\frac{7}{2}}\left(u\right)\right)+C $$

Substitute back {eq}u=5\theta. {/eq}

$$\displaystyle = \frac{1}{5}\left(-\frac{2}{3}\cos ^{\frac{3}{2}}\left(5\theta\right)+\frac{2}{7}\cos ^{\frac{7}{2}}\left(5\theta\right)\right)+C $$

Where C is constant.



B) We have to solve the integration of $$\displaystyle I = \int \frac{\cos^5 t}{\sqrt{\sin t}} \, \mathrm{d}t $$

Use the algebraic property.

$$\begin{align*} \displaystyle &= \int \frac{\cos ^4\left(t\right)\cos \left(t\right)}{\sqrt{\sin \left(t\right)}}dt\\ \displaystyle &= \int \frac{\left(\cos ^2\left(t\right)\right)^2\cos \left(t\right)}{\sqrt{\sin \left(t\right)}}dt \end{align*} $$

Use the trigonometric identity.

$$\displaystyle = \int \frac{\left(1-\sin ^2\left(t\right)\right)^2\cos \left(t\right)}{\sqrt{\sin \left(t\right)}}dt $$

Apply substitution for {eq}u=\sin \left(t\right) \Rightarrow du=\cos \left(t\right) dt. {/eq}

$$\displaystyle = \int \frac{\left(1-u^2\right)^2}{\sqrt{u}}du $$

Expand:

$$\displaystyle = \int \frac{1}{\sqrt{u}}-2u^{\frac{3}{2}}+u^{\frac{7}{2}}du $$

Apply the sum rule.

$$\displaystyle = \int \frac{1}{\sqrt{u}}du-\int \:2u^{\frac{3}{2}}du+\int u^{\frac{7}{2}}du $$

Use the exponent property and take the constant out.

$$\displaystyle = \int \:u^{-\frac{1}{2}}du-2\cdot \int u^{\frac{3}{2}}du+\int u^{\frac{7}{2}}du $$

Apply the power rule.

$$\displaystyle = 2u^{\frac{1}{2}}-\frac{4}{5}u^{\frac{5}{2}}+\frac{2}{9}u^{\frac{9}{2}}+C $$

Substitute back {eq}u=\sin \left(t\right). {/eq}

$$\displaystyle = 2\sqrt{\sin \left(t\right)}-\frac{4}{5}\sin ^{\frac{5}{2}}\left(t\right)+\frac{2}{9}\sin ^{\frac{9}{2}}\left(t\right)+C $$

Where C is constant.


Learn more about this topic:

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How to Solve Integrals Using Substitution

from Math 104: Calculus

Chapter 13 / Lesson 5
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