Evaluate the integral. a. \int 3 sin \sqrt {2t} dt b. \int_{-\infty}^0 \frac {2}{3x + 1} dx...

Question:

Evaluate the integral.

a. {eq}\int 3 sin \sqrt {2t} dt {/eq}

b. {eq}\int_{-\infty}^0 \frac {2}{3x + 1} dx {/eq}

c. {eq}\int_0^x \frac {-3 sin x}{\sqrt {cos x}} dx {/eq}

Integration by Substitution:

The integration which uses the u-substitution or v-substitution or w-substitution is called the integral by substitution method.

Historical important rules:

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

2. Integration by parts rule: {eq}\displaystyle \int p \frac{dq}{dy}dy= p q - \int \frac{dp}{dy}q dy. {/eq}

3. Common integration: {eq}\displaystyle \int \cos \left(u\right)du=\sin \left(u\right). {/eq}

4. Common integration: {eq}\displaystyle \int \frac{1}{u}du=\ln \left(\left|u\right|\right). {/eq}

5. The exponent property: {eq}\displaystyle \frac{1}{\sqrt{u}}=u^{-\frac{1}{2}}. {/eq}

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

Answer and Explanation:

a. We have to solve the integration of $$\displaystyle I = \int 3 sin \sqrt {2t} dt$$

Take the constant out.

$$\displaystyle = 3\cdot \int \sin \left(\sqrt{2t}\right)dt$$

Apply the substitution for {eq}u=\sqrt{2t} \Rightarrow du = \frac{1}{\sqrt{2t}} {/eq}

$$\displaystyle = 3\cdot \int u\sin \left(u\right)du$$

Apply integration by parts for {eq}u=u {/eq} and {eq}v'=\sin \left(u\right). {/eq}

$$\displaystyle = 3\left(-u\cos \left(u\right)-\int:-\cos \left(u\right)du\right)$$

Take the constant out.

$$\displaystyle = 3\left(-u\cos \left(u\right)-\left(-\int \cos \left(u\right)du\right)\right)$$

Use the common integration.

$$\displaystyle = 3\left(-u\cos \left(u\right)-\left(-\sin \left(u\right)\right)\right)+C$$

Substitute back {eq}u=\sqrt{2t}. {/eq}

$$\displaystyle = 3\left(-\sqrt{2t}\cos \left(\sqrt{2t}\right)-\left(-\sin \left(\sqrt{2t}\right)\right)\right)+C$$

Simplify:

$$\displaystyle = 3\left(-\sqrt{2t}\cos \left(\sqrt{2t}\right)+\sin \left(\sqrt{2t}\right)\right)+C$$

Where C is constant.

b. We have to solve the integration of $$\displaystyle I = \int_{-\infty}^0 \frac {2}{3x + 1} dx$$

This integral has undefined point at {eq}-\frac{1}{3}. {/eq}

So, integral can be weritten as:

$$\displaystyle = \int _{-\infty}^{-\frac{1}{3}}\frac{2}{3x+1}dx+\int _{-\frac{1}{3}}^0\frac{2}{3x+1}dx$$

Compute the indefinite integral.

$$\displaystyle = \int \frac{2}{3x+1}dx$$

Take the constant out.

$$\displaystyle = 2\cdot \int \frac{1}{3x+1}dx$$

Apply the substitution for {eq}u=3x+1 \Rightarrow du = 3 dx. {/eq}

$$\displaystyle = 2\cdot \int \frac{1}{3u}du$$

Take the constant out.

$$\displaystyle = 2\cdot \frac{1}{3}\cdot \int \frac{1}{u}du$$

Use the common integration.

$$\displaystyle = 2\cdot \frac{1}{3}\ln \left|u\right|+C$$

Substitute back {eq}u=3x+1. {/eq}

$$\displaystyle = 2\cdot \frac{1}{3}\ln \left|3x+1\right|+C$$

Simplify:

$$\displaystyle = \frac{2}{3}\ln \left|3x+1\right|+C$$

Where C is constant.

Now, compute the boundaries.

\begin{align*} \displaystyle &= -\infty -\infty\\ \displaystyle &= \mathrm{diverges}. \end{align*}

c. We have to solve the integration of $$\displaystyle I = \int_0^x \frac {-3 sin x}{\sqrt {cos x}} dx$$

Take the constant out.

$$\displaystyle = -3\cdot \int _0^x\frac{\sin \left(x\right)}{\sqrt{\cos \left(x\right)}}dx$$

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

Limits: {eq}0 rightarrow 1 {/eq} and {eq}x rightarrow \cos \left(x\right). {/eq}

$$\displaystyle = -3\cdot \int _1^{\cos \left(x\right)}-\frac{1}{\sqrt{u}}du$$

Take the constant out.

$$\displaystyle = -3\left(-\int _1^{\cos \left(x\right)}\frac{1}{\sqrt{u}}du\right)$$

Use the exponent property.

$$\displaystyle = -3\left(-\int _1^{\cos \left(x\right)}u^{-\frac{1}{2}}du\right)$$

Apply the power rule.

$$\displaystyle = 3\left[2u^{\frac{1}{2}}\right]^{\cos \left(x\right)}_1$$

Now, compute the boundaries.

$$\displaystyle = 3\left(2\cos ^{\frac{1}{2}}\left(x\right)-2\right).$$

Learn more about this topic:

How to Solve Integrals Using Substitution

from Math 104: Calculus

Chapter 13 / Lesson 5
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