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Integrals involving tan \(x\) and sec \(x\) Evaluate the following integrals. $$\int \tan ^{9} x \sec ^{4} x d x$$

Short Answer

Expert verified
Question: Evaluate the integral $\int \tan^{9} x \sec^{4} x d x$. Answer: The integral $\int \tan^{9} x \sec^{4} x d x = \frac{1}{9}(\sec^2 x - 1)^{\frac{9}{2}} + C$.

Step by step solution

01

Choose a substitution

Let us choose the substitution \(u = \sec x\). This means that \(du = \sec x \tan x dx\). We notice that if we differentiate \(\sec x\) with respect to \(x\), we get the \(\sec x \tan x\) term that is present in the integral.
02

Rewrite the integral in terms of \(u\)

To rewrite our integral, we need to express \(\tan^{9}x\), \(\sec^4x\) and \(dx\) in terms of \(u\). Since we have \(u = \sec x\), it's easy to write \(\sec^4x\) as \(u^4\). Now, let's express \(\tan^{9} x\) in terms of \(u\). We know that \(\tan^2 x = \sec^2 x - 1\), so \(\tan^9x = (\sec^2x - 1)^{\frac{9}{2}}\). Therefore, using our substitution, we have \(\tan^9x = (u^2 - 1)^{\frac{9}{2}}\). Finally, we rewrite \(dx\) using our substitution: \(dx = \frac{du}{\sec x \tan x} = \frac{du}{u(\sqrt{u^2 - 1})}\). Now let's substitute everything back into our integral: $$\int \tan^{9}x \sec^{4}x dx = \int(u^4)(u^2 - 1)^{\frac{9}{2}}\frac{du}{u(\sqrt{u^2 - 1})}$$
03

Simplify the integral and evaluate

After doing the substitution in Step 2, we can simplify the integral as follows: $$\int(u^4)(u^2 - 1)^{\frac{9}{2}}\frac{du}{u(\sqrt{u^2 - 1})} = \int u^3(u^2 - 1)^{\frac{9}{2}}\frac{du}{\sqrt{u^2 - 1}}$$ Now, let's make another substitution: \(v = u^2 - 1\). Then, \(dv = 2u du\). Since our integral involves \(u^3 du\), let us multiply and divide by \(2\), so our integral becomes: $$\frac{1}{2}\int \frac{u^3 du}{\sqrt{u^2 - 1}} = \frac{1}{2}\int \frac{(2u)u^2 du}{\sqrt{u^2 - 1}}$$ Now we can rewrite the integral in terms of \(v\): $$\frac{1}{2}\int \frac{v^{\frac{9}{2}}}{\sqrt{v}}dv = \frac{1}{2}\int v^{\frac{7}{2}}dv$$ Now, we can integrate this expression: $$\frac{1}{2}\int v^{\frac{7}{2}}dv = \frac{1}{2} \cdot \frac{2}{9}v^{\frac{9}{2}} + C$$
04

Substitute back the original variables

Now, we substitute \(v\) and \(u\) back into the expression to obtain the solution in terms of \(x\): $$\frac{1}{9}(u^2 - 1)^{\frac{9}{2}} = \frac{1}{9}(\sec^2 x - 1)^{\frac{9}{2}} + C$$ Finally, the evaluated integral is: $$\int \tan^{9} x \sec^{4} x d x = \frac{1}{9}(\sec^2 x - 1)^{\frac{9}{2}} + C$$

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Key Concepts

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

Trigonometric Substitution
Trigonometric substitution is a powerful tool in calculus, especially when dealing with integrals involving radical expressions, where direct methods may not be obvious. It leverages the Pythagorean identities of trigonometry to simplify the integration process.

When faced with an integral that includes expressions like \(\sqrt{a^2 - x^2}\), \(\sqrt{a^2 + x^2}\), or \(\sqrt{x^2 - a^2}\), trigonometric substitution suggests replacing \(x\) with a trigonometric function of a new variable, usually \(\theta\), that corresponds to one side of a right triangle.

For instance, in the integral \(\int \tan^{9} x \sec^{4} x dx\), the substitution \(u = \sec x\) is chosen due to its derivative being a product of secant and tangent, which are present in the integral. This clever selection simplifies the given expression significantly, allowing for further steps in the solution.
Integration Techniques
When integrating complex functions, several techniques can be applied to reach the solution. These include methods like substitution, integration by parts, partial fractions, and trigonometric substitution, as mentioned earlier.

Choosing the right technique is often the first major step toward solving an integral. For example, the integral \(\int \tan^{9} x \sec^{4} x dx\) initially seems daunting. By identifying an appropriate substitution, such as \(u = \sec x\), the integral becomes much more manageable.

Integration techniques are often used in combination. After performing a trigonometric substitution, one might further simplify the resulting expression by applying u-substitution a second time, exploiting derivative relations that surface, like substituting \(v = u^2 - 1\) after first substituting \(u = \sec x\).
U-Substitution
U-substitution, also known as integration by substitution, is a method that involves changing the variable of integration to simplify the integral. The substitution is typically a function \(u\) that is chosen to either simplify the integrand or to match the form of an integrable function.

Consider the integral \(\int \tan^{9} x \sec^{4} x dx\). After identifying \(u = \sec x\), which simplifies \(\sec^4 x\) to \(u^4\), the next step is to rewrite the entire integrand in terms of \(u\), incorporating \(du\) to replace \(dx\). The key to a successful u-substitution is recognizing which part of the integral to transform, and formulating the change of variables in a way that the new integral is simpler or familiar.

It's essential to express \(dx\) in terms of \(du\) correctly. In this example, after a clever manipulation, \(dx\) becomes \(\frac{du}{u(\sqrt{u^2 - 1})}\), further aiding the simplification of the original integral.
Integration by Substitution
Integration by substitution is often used interchangeably with u-substitution, but it is the broader concept of changing variables to make the integral more manageable. This method is a cornerstone of integral calculus and can be likened to the reverse process of the chain rule for differentiation.

In the exercise \(\int \tan^{9} x \sec^{4} x dx\), integration by substitution is not only used once but twice. Initially, \(u = \sec x\) is substituted to rewrite the integral in terms of \(u\). Subsequently, another substitution, \(v = u^2 - 1\), is made to reduce the integrand to a form that is readily integrable. By doing this, we deal with an integral of the form \(\int v^{\frac{7}{2}}dv\), which is straightforward to evaluate.

Once the integration is performed in the \(v\) variable, the final step involves back-substituting the original variables to express the solution in terms of the original variable \(x\), yielding the final integrated function.

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