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$$\text { Evaluate the following integrals.}$$ $$\int 20 \tan ^{6} x d x$$

Short Answer

Expert verified
Short Answer: The integral of \(20 \tan^6 x dx\) is equal to C, where C is the constant of integration.

Step by step solution

01

Recall the common integral formula for tangent

Recall the integral formula for the tangent function: $$\int \tan^n x dx = \int (\sin x)^n(\cos x)^{-n} dx$$
02

Apply the common integral formula to the given exercise

Apply the formula to the given integral: $$\int 20\tan^6 x dx = 20 \int (\sin x)^6 (\cos x)^{-6} dx$$
03

Simplify the integral and perform the substitution

Now we perform the substitution \(u = \sin x\), which implies \(du = \cos x dx\): $$20\int (\sin x)^6 (\cos x)^{-6} dx = 20\int u^6 (\cos x)^{-6} (\cos x dx)$$ $$= 20\int u^6 (\cos x)^{-6} du$$
04

Rewrite the integral using the secant function

We can rewrite the integral using the secant function since \(\sec x = \frac{1}{\cos x}\): $$20\int u^6 (\cos x)^{-6} du = 20\int u^6 (\sec x)^6 du$$
05

Perform the substitution back to tangent

Now substitute back the tangent function: $$u = \sin x = \tan x \cdot \frac{1}{\sec x} = \tan x \cos x$$ $$20\int u^6 (\sec x)^6 du = 20\int (\tan x \cos x)^6 (\sec x)^6 du = 20\int (\tan x)^6 (\cos x)^{-6} du$$
06

Rewrite the integral with respect to x

Now rewrite the integral with respect to \(x\): $$20\int (\tan x)^6 (\cos x)^{-6} du = 20\int (\tan x)^6 (\cos x)^{-6} dx$$
07

Evaluate the final integral

Finally, evaluate the integral: $$\int 20 \tan ^{6} x d x = 20\int (\tan x)^6 (\cos x)^{-6} dx = C$$ The integral is \(\int 20 \tan ^{6} x d x = C\), where C is the constant of integration.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometric Integrals
Trigonometric integrals involve integrating expressions containing trigonometric functions like sine, cosine, and tangent. When dealing with these integrals, special techniques are often required due to the periodic nature and identities associated with trigonometric functions.

A common method to address trigonometric integrals is by using trigonometric identities that simplify the expressions. For example, in the original exercise, the task was to evaluate \( \int 20 \tan^6 x \ dx \). By expressing the tangent function as \( \tan x = \frac{\sin x}{\cos x} \), we rewrite the integral in terms of sine and cosine.
  • This can simplify integration by leveraging known integral forms and trigonometric identities.
  • It is helpful to remember key identities like \( \sin^2 x + \cos^2 x = 1 \) which can be substituted to simplify integrals.
Note that different substitutions and identities can dramatically change the complexity of the resulting integral.
Substitution Method
The substitution method is a powerful technique in integral calculus used to simplify complex integrals. Essentially, substitution involves changing the variable in the integral to a function of a new variable that makes the integral easier to evaluate.

In the given problem, substitution is initiated by letting \( u = \sin x \), leading to \( du = \cos x \, dx \). This technique transforms the integral into a simpler form, often related to a function that's easier to integrate.
  • Identify a variable to substitute that transforms difficult integrals into standard forms.
  • Ensure to change the differential as well, from \( dx \) to \( du \) in this case.
  • Don't forget to convert back to the original variable, ensuring the solution is expressed correctly.
Effective substitution often involves recognizing patterns and can simplify an integral, making previously challenging problems approachable.
Integral Calculus
Integral calculus is the study of integrals and their properties, focusing particularly on accumulating quantities and finding areas under curves. At its core, integral calculus seeks to reverse the process of differentiation. Trigonometric integrals are a subset of this study that often require unique approaches.

An integral, such as \( \int 20 \tan^6 x \ dx \), involves evaluating how a trigonometric function accumulates over an interval. The result is an expression plus an arbitrary constant \( C \), representing the family of antiderivatives. In this context:
  • The definite integral gives the net area between a function and the x-axis over an interval.
  • The indefinite integral provides a general form with a constant due to the lack of bounds.
  • Understanding properties like linearity of integrals aids in the manipulation and simplification of complex integrals.
Integral calculus is a fundamental part of mathematical analysis and its applications are vast, spanning from physics to economics, where it models and predicts continuous change.

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Most popular questions from this chapter

Let \(R\) be the region between the curves \(y=e^{-c x}\) and \(y=-e^{-c x}\) on the interval \([a, \infty),\) where \(a \geq 0\) and \(c \geq 0 .\) The center of mass of \(R\) is located at \((\bar{x}, 0)\) where \(\bar{x}=\frac{\int_{a}^{\infty} x e^{-c x} d x}{\int_{a}^{\infty} e^{-c x} d x} .\) (The profile of the Eiffel Tower is modeled by the two exponential curves.) a. For \(a=0\) and \(c=2,\) sketch the curves that define \(R\) and find the center of mass of \(R\). Indicate the location of the center of mass. b. With \(a=0\) and \(c=2,\) find equations of the lines tangent to the curves at the points corresponding to \(x=0.\) c. Show that the tangent lines intersect at the center of mass. d. Show that this same property holds for any \(a \geq 0\) and any \(c>0 ;\) that is, the tangent lines to the curves \(y=\pm e^{-c x}\) at \(x=a\) intersect at the center of mass of \(R\) (Source: P. Weidman and I. Pinelis, Comptes Rendu, Mechanique \(332(2004): 571-584 .)\)

Find the volume of the described solid of revolution or state that it does not exist. The region bounded by \(f(x)=(x+1)^{-3 / 2}\) and the \(y\) -axis on the interval (-1,1] is revolved about the line \(x=-1.\)

Recall that the substitution \(x=a \sec \theta\) implies that \(x \geq a\) (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 \(\tan \theta \leq 0\) ). $$ \begin{array}{l} \text { Show that } \int \frac{d x}{x \sqrt{x^{2}-1}}= \\ \qquad\left\\{\begin{array}{ll} \sec ^{-1} x+C=\tan ^{-1} \sqrt{x^{2}-1}+C & \text { if } x>1 \\ -\sec ^{-1} x+C=-\tan ^{-1} \sqrt{x^{2}-1}+C & \text { if } x<-1 \end{array}\right. \end{array} $$

An integrand with trigonometric functions in the numerator and denominator can often be converted to a rational integrand using the substitution \(u=\tan (x / 2)\) or \(x=2 \tan ^{-1} u .\) The following relations are used in making this change of variables. $$A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}$$ $$\text { Evaluate } \int \frac{d x}{1-\cos x}$$

Use symmetry to evaluate the following integrals. a. \(\int_{-\infty}^{\infty} e^{|x|} d x \quad\) b. \(\int_{-\infty}^{\infty} \frac{x^{3}}{1+x^{8}} d x\)

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