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Chapter 7: Problem 36

Evaluate the following definite integrals. $$\int_{0}^{1 / \sqrt{2}} y \tan ^{-1} y^{2} d y$$

Short Answer

Expert verified
Question: Evaluate the definite integral of the function \(y\tan^{-1}(y^2)\) over the interval \([0, \frac{1}{\sqrt{2}}]\). Answer: The definite integral evaluates to \(\frac{1}{4}\tan^{-1}\left(\frac{1}{2}\right) - \frac{1}{4}\ln(2)\).

Step by step solution

01

Choose a substitution

We have the function \(y\tan^{-1}(y^2)\), and we want to find the integral of this function. We choose the substitution \(u = y^2\). This allows us to transform the function into a more manageable form.
02

Calculate the differential

Differentiating with respect to \(y\), we obtain \(du = 2y\, dy\). Note that we want \(\int y\,dy\), so changing into \(y\) gives us \(y\,dy = \frac{1}{2}du\).
03

Update the limits of integration

Since we changed the variable from \(y\) to \(u\), the interval of integration must also be changed. When \(y=0\), we have \(u = 0^2 = 0\). When \(y = \frac{1}{\sqrt{2}}\), we have \(u = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}\). Our new interval of integration is \([0, \frac{1}{2}]\).
04

Rewrite the integral in terms of new variable

Now we rewrite the integral in terms of the new variable \(u\). The original integral becomes: $$\int_{0}^{\frac{1}{2}} \left(\frac{1}{2} \tan^{-1}u\right) du$$
05

Integrate the new function

Now we integrate the new function with respect to \(u\): $$\frac{1}{2} \int_{0}^{\frac{1}{2}} \tan^{-1}u\, du$$ At this point, we can use the integration by parts method to evaluate the integral, where we let \(v' = 1\) and \(u' = \tan^{-1}(u)\).
06

Apply integration by parts

Using the integration by parts formula, which is \(\int u'v = uv - \int v'u\), and after letting \(v' = 1\) and \(u' = \tan^{-1}(u)\), we derive \(v = u\) and \(u = \tan^{-1}(u)\). Applying the formula, we get: $$\frac{1}{2}\left[u\tan^{-1}(u)\biggr|_0^{\frac{1}{2}} - \int_{0}^{\frac{1}{2}} u\frac{1}{1 + u^2} du\right]$$
07

Evaluate the first term and simplify the integral

Evaluating the first term of the expression at the limits of integration: $$\frac{1}{2}\left[\left(\frac{1}{2}\tan^{-1}\left(\frac{1}{2}\right)-0\right)\right] = \frac{1}{4}\tan^{-1}\left(\frac{1}{2}\right)$$ Now, we simplify the integral in the expression: $$-\frac{1}{2}\int_{0}^{\frac{1}{2}} \frac{u}{1 + u^2} du$$
08

Integrate the remaining term

We can perform this integral using another substitution. Let \(w = 1 + u^2\) so that \(dw = 2u\, du\). The integral becomes: $$\frac{1}{2}\int_{0}^{\frac{1}{2}} \frac{u}{1 + u^2} du = -\frac{1}{4}\int_1^2 \frac{1}{w} dw$$ Now, integrate the term: $$-\frac{1}{4}\left[\ln(w)\biggr|_1^2\right] = -\frac{1}{4}\ln(2)$$
09

Combine results

Finally, we combine the results of both terms together to get the result: $$\frac{1}{4}\tan^{-1}\left(\frac{1}{2}\right) - \frac{1}{4}\ln(2)$$

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Most popular questions from this chapter

Evaluate \(\int \frac{d x}{x^{2}-1},\) for \(x > 1,\) in two ways: using partial fractions and a trigonometric substitution. Reconcile your two answers.

Compute \(\int_{0}^{1} \ln x d x\) using integration by parts. Then explain why \(-\int_{0}^{\infty} e^{-x} d x\) (an easier integral) gives the same result.

When is the volume finite? Let \(R\) be the region bounded by the graph of \(f(x)=x^{-p}\) and the \(x\) -axis, for \(x \geq 1.\) a. Let \(S\) be the solid generated when \(R\) is revolved about the \(x\) -axis. For what values of \(p\) is the volume of \(S\) finite? b. Let \(S\) be the solid generated when \(R\) is revolved about the \(y\) -axis. For what values of \(p\) is the volume of \(S\) finite?

For what values of \(p\) does the integral \(\int_{2}^{\infty} \frac{d x}{x \ln ^{p} x}\) exist and what is its value (in terms of \(p\) )?

a. Verify the identity \(\sec x=\frac{\cos x}{1-\sin ^{2} x}\) b. Use the identity in part (a) to verify that \(\int \sec x d x=\frac{1}{2} \ln \left|\frac{1+\sin x}{1-\sin x}\right|+C\) (Source: The College Mathematics Joumal \(32,\) No. 5 (November 2001))
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