Evaluate the integral of integral z^5(4-z^2)^{fraction {5}{2} } dz from fraction {1}{2} to 1

Question:

Evaluate the integral of {eq}\displaystyle \int z^5(4-z^2)^{\frac {5}{2} } dz {/eq} from {eq}\frac {1}{2} \ \ to \ \ 1 {/eq}

Integration by Trigonometric Substitution:

The use of trigonometric function in the integration is known as the trigonometric substitution method. For example, the radical expressions which cannot solve directly, then we use the trigonometric functions as a trigonometric substitution which can easily solve by simple integration method.

Historical context used for this integration:

1. Eject the constant out: {eq}\displaystyle \int l\cdot f\left(z\right)dz=l\cdot \int f\left(z\right)dz. {/eq}

2. Common derivative: {eq}\displaystyle \frac{d}{dx}\left(\sin \left(x\right)\right)=\cos \left(x\right). {/eq}

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

4. Common derivative: {eq}\displaystyle \frac{d}{dx}\left(\cos \left(x\right)\right)=-\sin \left(x\right). {/eq}

5. The sum rule: {eq}\displaystyle \int p\left(z\right)\pm q\left(z\right)dz=\int p\left(z\right)dz\pm \int q\left(z\right)dz. {/eq}

6. The power rule: {eq}\displaystyle \int z^l dz=\frac{z^{l+1}}{l+1}, \quad l\ne -1. {/eq}

Answer and Explanation:

We have to evaluate the integration of $$\displaystyle I = \int z^5(4-z^2)^{\frac {5}{2} } dz $$

Apply trigonometric substitution for {eq}z=2\sin \left(u\right) \Rightarrow dz = 2\cos \left(u\right) du. {/eq}

Limits: {eq}\frac{1}{2} \rightarrow \arcsin \left(\frac{1}{4}\right) {/eq} and {eq}1 \rightarrow \frac{\pi }{6}. {/eq}

$$\displaystyle = \int _{\arcsin \left(\frac{1}{4}\right)}^{\frac{\pi }{6}}2048\cos ^6\left(u\right)\sin ^5\left(u\right)du $$

Eject the constant out.

$$\displaystyle = 2048\cdot \int _{\arcsin \left(\frac{1}{4}\right)}^{\frac{\pi }{6}}\cos ^6\left(u\right)\sin ^5\left(u\right)du $$

Use the algebraic property.

$$\begin{align*} \displaystyle &= 2048\cdot \int _{\arcsin \left(\frac{1}{4}\right)}^{\frac{\pi }{6}}\cos ^6\left(u\right)\sin ^4\left(u\right)\sin \left(u\right)du\\ \displaystyle &= 2048\cdot \int _{\arcsin \left(\frac{1}{4}\right)}^{\frac{\pi }{6}}\cos ^6\left(u\right)\left(\sin ^2\left(u\right)\right)^2\sin \left(u\right)du \end{align*} $$

Use the trigonometric identity.

$$\displaystyle = 2048\cdot \int _{\arcsin \left(\frac{1}{4}\right)}^{\frac{\pi }{6}}\cos ^6\left(u\right)\left(1-\cos ^2\left(u\right)\right)^2\sin \left(u\right)du $$

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

Limits: {eq}\arcsin \left(\frac{1}{4}\right) \rightarrow \frac{15^{\frac{1}{2}}}{4} {/eq} and {eq}\frac{\pi }{6} \rightarrow \frac{\sqrt{3}}{2}. {/eq}

$$\displaystyle = 2048\cdot \int _{\frac{15^{\frac{1}{2}}}{4}}^{\frac{\sqrt{3}}{2}}-v^6\left(1-v^2\right)^2dv $$

Expand:

$$\displaystyle = 2048\cdot \int _{\frac{15^{\frac{1}{2}}}{4}}^{\frac{\sqrt{3}}{2}}-v^6+2v^8-v^{10}dv $$

Change the limits.

$$\displaystyle = 2048\left(-\int _{\frac{\sqrt{3}}{2}}^{\frac{15^{\frac{1}{2}}}{4}}-v^6+2v^8-v^{10}dv\right) $$

Apply the sum rule.

$$\displaystyle = 2048\left(-\left(-\int _{\frac{\sqrt{3}}{2}}^{\frac{15^{\frac{1}{2}}}{4}}v^6dv+\int _{\frac{\sqrt{3}}{2}}^{\frac{15^{\frac{1}{2}}}{4}}2v^8dv-\int _{\frac{\sqrt{3}}{2}}^{\frac{15^{\frac{1}{2}}}{4}}v^{10}dv\right)\right) $$

Eject the constant out.

$$\displaystyle = 2048\left(-\left(-\int _{\frac{\sqrt{3}}{2}}^{\frac{15^{\frac{1}{2}}}{4}}v^6dv+2\cdot \int _{\frac{\sqrt{3}}{2}}^{\frac{15^{\frac{1}{2}}}{4}}v^8dv-\int _{\frac{\sqrt{3}}{2}}^{\frac{15^{\frac{1}{2}}}{4}}v^{10}dv\right)\right) $$

Apply the power rule.

$$\displaystyle = 2048\left(-\left(-\left[\frac{v^7}{7}\right]^{\frac{15^{\frac{1}{2}}}{4}}_{\frac{\sqrt{3}}{2}}+2\left[\frac{v^9}{9}\right]^{\frac{15^{\frac{1}{2}}}{4}}_{\frac{\sqrt{3}}{2}}-\left[\frac{v^{11}}{11}\right]^{\frac{15^{\frac{1}{2}}}{4}}_{\frac{\sqrt{3}}{2}}\right)\right) $$

Compute the boundaries.

$$\begin{align*} \displaystyle &= 2048\left(-\left(-\left[\frac{3375\cdot 15^{\frac{1}{2}}}{114688}-\frac{27\cdot \:3^{\frac{1}{2}}}{896}\right]+2\left[\frac{5625\cdot 15^{\frac{1}{2}}}{262144}-\frac{9\cdot 3^{\frac{1}{2}}}{512}\right]-\left[\frac{759375\cdot 15^{\frac{1}{2}}}{46137344}-\frac{243\cdot 3^{\frac{1}{2}}}{22528}\right]\right)\right)\\ \displaystyle &= -2048\cdot \frac{1861632\cdot \:3^{\frac{1}{2}}-959625\cdot \:15^{\frac{1}{2}}}{322961408}\\ \displaystyle &= -\frac{-959625\cdot \:15^{\frac{1}{2}}+1861632\cdot \:3^{\frac{1}{2}}}{157696}. \end{align*} $$


Learn more about this topic:

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How to Use Trigonometric Substitution to Solve Integrals

from Math 104: Calculus

Chapter 13 / Lesson 11
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