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1) Find the derivative of the function. y = ?^s i n x_ c o s x l n ( 8 + 6 v ) d v 2) Find...

Question:

1) Find the derivative of the function. {eq}y=\int_{cos \ x}^{sin \ x}ln(8+6v)dv {/eq}
2) Find the general indefinite integral. {eq}\int \left ( 2x^{2}+3x^{-2} \right )dx {/eq}
3) Find the general indefinite integral.{eq}\int 5v(v^{2}+8)^{2} dv {/eq}

Insert context header here:

There are many techniques in evaluating integrals. Two of the most common are power rule and substitution. In this problem, we evaluated two indefinite integrals and one definite integral

Answer and Explanation:

a. This integral can be evaluated by substitution. We let {eq}u=8+6v{/eq} and {eq}du=6dv{/eq}. So the integral becomes:

$$y=\int\ln(8+6v)dv=\frac{1}{6}\int\ln{udu}=\frac{1}{6}(u\ln{u}-u)$$

We revert to the original variable by doing the same substitutions and evaluate the definite integral:

$$y=\frac{1}{6}({(8+6v)\ln(8+6v)}-(8+6v))|_{\cos{x}}^{\sin{x}}\\ y=\frac{1}{6}({(8+6\sin{x})\ln(8+6\sin{x})}-(8+6\sin{x}))-\frac{1}{6}({(8+6\cos{x})\ln(8+6\cos{x})}-(8+6\cos{x}))\\ $$

Applying the properties of natural logarithm:

$$y=\frac{1}{6}\ln\frac{(8+6\sin{x})^{8+6\sin{x}}}{(8+6\cos{x})^{8+6\cos{x}}}+\cos{x}-\sin{x}$$

b. The second integral was evaluated using power rule:

$$\int (2x^2+3x^{-1})dx=\frac{2x^3}{3}+\frac{3x^{-1}}{-1}=\frac{2x^3}{3}-3x^{-1}+C$$

c. This integral was evaluated using substitution. We let {eq}u=v^2+8{/eq} and {eq}du=2vdv:{/eq}

$$\int 5v(v^2+8)dv=\frac{5}{2}\int{u^2}du=\frac{5}{2}\cdot\frac{u^3}{3}+C $$

We revert to the original variable by doing the same substitution:

$$\frac{5}{6}(v^2+8)^3+C $$


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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