/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 22

Evaluate the following integrals. $$\int \frac{d y}{y^{-1}+y^{-3}}$$

Short Answer

Expert verified
Question: Evaluate the integral of the following expression: $$\int \frac{dy}{y^{-1}+y^{-3}}$$ Answer: The integral of the given expression can be evaluated as: $$\int \frac{dy}{y^{-1}+y^{-3}} = \frac{1}{2} \left(\frac{1}{2}(y^2+1)^2 - (y^2+1)\right) + C$$

Step by step solution

01

Simplify the Integrands

In this step, we will simplify the expression inside the integral. $$\int \frac{dy}{y^{-1}+y^{-3}} = \int \frac{dy}{\frac{1}{y}+\frac{1}{y^3}} $$ Now, we will combine the terms in the denominator by finding a common denominator: $$\int \frac{dy}{\frac{1}{y}+\frac{1}{y^3}} = \int \frac{dy}{\frac{y^2+1}{y^3}} $$ Now, we can simplify the fraction: $$\int \frac{dy}{\frac{y^2+1}{y^3}} = \int y^3dy$$ Step 2: Perform a substitution
02

Substitution

We perform the substitution $$u=y^2+1$$Then, we find the derivative of u with respect to y: $$\frac{du}{dy} = 2y$$Now, let's rewrite our integral in terms of u: $$\int y^3dy = \int \frac{y^2}{2} \frac{du}{dy}dy$$Substitute u for \(y^2\) in the integral:$$\int \frac{y^2}{2} \frac{du}{dy}dy = \int \frac{u-1}{2} \frac{du}{dy}dy$$Now, substitute du for \(\frac{du}{dy}dy\):$$\int \frac{u-1}{2} \frac{du}{dy}dy = \int \frac{u-1}{2} du$$ Step 3: Integrate with respect to u
03

Integration

Now we can proceed to integrate the expression with respect to u: $$\int \frac{u-1}{2} du =\frac{1}{2}\int (u-1) du $$Now, we will integrate term by term:$$\frac{1}{2}\int (u-1) du = \frac{1}{2} \left(\int u \ du - \int 1 \ du\right)$$After integrating both terms, we get:$$\frac{1}{2} \left(\int u \ du - \int 1 \ du\right) = \frac{1}{2} \left(\frac{1}{2}u^2 - u\right) + C$$where C is the constant of integration. Step 4: Convert back to y
04

Back Substitution

Finally, we will convert our result back into terms of y using our substitution:$$\frac{1}{2} \left(\frac{1}{2}u^2 - u\right) + C = \frac{1}{2} \left(\frac{1}{2}(y^2+1)^2 - (y^2+1)\right) + C$$This is the final result of our integral: $$\int \frac{dy}{y^{-1}+y^{-3}} = \frac{1}{2} \left(\frac{1}{2}(y^2+1)^2 - (y^2+1)\right) + C$$

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Most popular questions from this chapter

Recall that the substitution \(x=a \sec \theta\) implies either \(x \geq a\) (in which case \(0 \leq \theta<\pi / 2\) and \(\tan \theta \geq 0)\) or \(x \leq-a\) (in which case \(\pi / 2<\theta \leq \pi\) and \(\tan \theta \leq 0\) ). Graph the function \(f(x)=\frac{\sqrt{x^{2}-9}}{x}\) and consider the region bounded by the curve and the \(x\) -axis on \([-6,-3] .\) Then evaluate \(\int_{-6}^{-3} \frac{\sqrt{x^{2}-9}}{x} d x .\) Be sure the result is consistent with the graph.

Evaluating an integral without the Fundamental Theorem of Calculus Evaluate \(\int_{0}^{\pi / 4} \ln (1+\tan x) d x\) using the following steps. a. If \(f\) is integrable on \([0, b],\) use substitution to show that $$ \int_{0}^{b} f(x) d x=\int_{0}^{b / 2}(f(x)+f(b-x)) d x $$ b. Use part (a) and the identity \(\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}\) to evaluate \(\int_{0}^{\pi / 4} \ln (1+\tan x) d x\) (Source: The College Mathematics Journal 33, 4, Sep 2004).

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