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Chapter 7: Problem 31

Evaluate the following integrals. $$\int 10 \tan ^{9} x \sec ^{2} x d x$$

Short Answer

Expert verified
The result of evaluating the integral \(\int 10 \tan ^{9} x \sec ^{2} x d x\) is \((\tan x)^{10} + C\), where C is the constant of integration.

Step by step solution

01

Identify the substitution

Notice that the derivative of the tangent function is the secant squared function: $$\frac{d}{dx}(\tan x) = \sec^2 x$$ So, we can make the substitution: $$u = \tan x$$ Then, the derivative of u with respect to x is: $$\frac{du}{dx} = \sec^2 x$$ And, rearranging to make dx the subject: $$dx=\frac{du}{\sec^2x}$$
02

Perform the substitution

Now that we have our substitution and the expression for dx, we can substitute them into the integral and simplify. $$\int 10 \tan ^{9} x \sec ^{2} x d x = \int 10 u^9 \sec^2x \frac{du}{\sec^2x}$$ The secant squared terms cancel out: $$\int 10 u^9 \sec^2x \frac{du}{\sec^2x} = \int 10 u^9 du$$
03

Evaluate the integral in terms of u

Now, integrate the expression with respect to u: $$\int 10 u^9 du = 10 \int u^9 du = 10 \frac{u^{10}}{10} + C$$ Simplifying, we get: $$u^{10} + C$$
04

Substitute back in terms of x

Finally, we need to substitute back in terms of x. From Step 1, we know that \(u = \tan x\), so we substitute this back into the expression we obtained in Step 3: $$u^{10} + C = (\tan x)^{10} + C$$ This is the final result: $$\int 10 \tan ^{9} x \sec ^{2} x d x = (\tan x)^{10} + C$$

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Most popular questions from this chapter

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