/*! 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 21 Find or evaluate the integral. ... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 21

Find or evaluate the integral. $$ \int x \tan ^{2} x d x $$

Short Answer

Expert verified
The integral of \( x \tan^{2} x \, dx \) is \( x(\tan{x} - x) + \ln{|\cos{x}|} + \frac{1}{2} x^2 - Cx + C \).

Step by step solution

01

Choose u and dv

Choose \(u = x\) and \(dv = \tan^2{x} dx\). We will need to find \(du\) and \(v\) next.
02

Find du and v

Differentiate \(u\) with respect to \(x\) to find \(du\): $$ du = \frac{d}{dx}(x) dx = dx $$ To find \(v\), integrate \(dv\) with respect to \(x\): $$ v = \int \tan^2{x} dx $$ Now, use the identity \(\tan^2{x} = \sec^2{x} - 1\) to rewrite the integral: $$ v = \int (\sec^2{x} - 1) dx $$ Separate the integral: $$ v = \int \sec^2{x} dx - \int 1 dx $$ Now, integrate each term. The first integral is a standard integral, and the second is a basic power function: $$ v = \tan{x} - x + C $$
03

Apply integration by parts formula

Now that we have \(u\), \(du\), and \(v\), we can apply the integration by parts formula: $$ \int x \tan^2{x} dx = x (\tan{x} - x + C) - \int (\tan{x} - x + C) dx $$
04

Integrate remaining integral

We need to integrate the remaining integral in the above expression: $$ \int (\tan{x} - x + C) dx = \int \tan{x} dx - \int x dx + \int C dx $$ Integrating each individual term, we get: $$ - \ln{|\cos{x}|} - \frac{1}{2} x^2 + Cx + D $$ Where D is another constant of integration.
05

Combine results

Combine both parts mentioned in step 3 and step 4: $$ \int x \tan^2{x} dx = x(\tan{x} - x) - (-\ln{|\cos{x}|} - \frac{1}{2} x^2 + Cx) + E $$ Where \(E\) is the combined constant of integration, which can be written simply as \(C\) again.
06

Final Answer

The final result for the integral we were given is: $$ \int x \tan^2{x} dx = x(\tan{x} - x) + \ln{|\cos{x}|} + \frac{1}{2} x^2 - Cx + C $$

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Most popular questions from this chapter

Let \(I=\int \frac{8 x^{5}-3 x^{4}+2 x^{2}-1}{36 x^{6}-108 x^{5}+105 x^{4}-72 x^{3}+58 x^{2}-12 x+9} d x\) a. Use a CAS to find the partial fraction decomposition of \(f(x)=\frac{8 x^{5}-3 x^{4}+2 x^{2}-1}{36 x^{6}-108 x^{5}+105 x^{4}-72 x^{3}+58 x^{2}-12 x+9}\) b. Use a CAS to find \(I\).

If \(f\) is defined on \([0, \infty)\), then the average value of \(f\) over \([0, \infty)\) is defined to be $$ f_{\mathrm{av}}=\lim _{b \rightarrow \infty} \frac{1}{b} \int_{0}^{b} f(x) d x $$ Suppose that the voltage and current in an AC circuit are $$ V=V_{0} \cos \omega t \quad \text { and } \quad I=I_{0} \cos (\omega t+\phi) $$ so that the voltage and current differ by an angle \(\phi\). Then the power output is \(P=V I\). Show that the average power output is \(P_{\mathrm{av}}=\frac{1}{2} I_{0} V_{0} \cos \phi\). Note: The factor \(\cos \phi\) is called the power factor. When \(V\) and \(I\) are in phase \(\left(\phi=0^{\circ}\right)\), the average power output is \(\frac{1}{2} I V\), but when \(V\) and \(I\) are out of phase \(\left(\phi=90^{\circ}\right)\), then the average power output is zero.

Let \(f(t)\) be continuous for \(t>0 .\) The Laplace transform of \(f\) is the function \(F\) defined by $$ F(s)=\int_{0}^{\infty} f(t) e^{-s t} d t $$ provided that the integral exists use this definition. Show that the Laplace transform of \(f(t)=\cos \omega t\) is \(F(s)=\frac{s}{s^{2}+\omega^{2}}\).

Observe that \(\int_{0}^{\infty} e^{-x^{2}} d x=\int_{0}^{4} e^{-x^{2}} d x+\int_{4}^{\infty} e^{-x^{2}} d x\). a. Show that \(\int_{4}^{\infty} e^{-x^{2}} d x \leq 10^{-7}\) so that \(\int_{0}^{\infty} e^{-x^{2}} d x \approx \int_{0}^{4} e^{-x^{2}} d x\) b. Use a calculator or computer to obtain an estimate for $\int_{0}^{\infty} e^{-x^{2}} d x.

Find or evaluate the integral. \(\int \frac{x}{\sqrt[3]{2-x}} d x\)
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