Evaluate $\int \cos(2x) \sqrt{1+\sin^2(2x)} \, Dx$.Hint: You May Use That $\int \sec^3(x) \, Dx = \frac{1}{2}(\sec(x) \tan(x) + \ln|\sec(x) + \tan(x)|) + C$.

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Introduction

In this article, we will evaluate the given integral $\int \cos(2x) \sqrt{1+\sin^2(2x)} \, dx$. This integral appears to be complex and may require some advanced techniques to solve. We will use the given hint to simplify the integral and eventually find its solution.

Step 1: Simplify the Integral

To simplify the integral, we can start by using the given hint that $\int \sec^3(x) \, dx = \frac{1}{2}(\sec(x) \tan(x) + \ln|\sec(x) + \tan(x)|) + C$. We can rewrite the integral as $\int \cos(2x) \sqrt{1+\sin^2(2x)} \, dx = \int \cos(2x) \sqrt{1+\sin^2(2x)} \cdot \frac{\cos(2x)}{\cos(2x)} \, dx$.

Step 2: Use Trigonometric Identity

We can use the trigonometric identity $\sin^2(2x) = 1 - \cos^2(2x)$ to rewrite the integral as $\int \cos(2x) \sqrt{2 - \cos^2(2x)} \, dx$. This simplifies the integral and makes it easier to work with.

Step 3: Substitute $u = \cos(2x)$

We can substitute $u = \cos(2x)$ to simplify the integral further. This gives us $\int \sqrt{2 - u^2} \, du$. We can now use the given hint to evaluate this integral.

Step 4: Evaluate the Integral

Using the given hint, we can evaluate the integral as $\int \sqrt{2 - u^2} \, du = \frac{1}{2}u\sqrt{2-u^2} + \frac{1}{2}\sin^{-1}\left(\frac{u}{\sqrt{2}}\right) + C$. We can now substitute back $u = \cos(2x)$ to find the final solution.

Step 5: Substitute Back $u = \cos(2x)$

Substituting back $u = \cos(2x)$, we get $\frac{1}{2}\cos(2x)\sqrt{2-\cos^2(2x)} + \frac{1}{2}\sin^{-1}\left(\frac{\cos(2x)}{\sqrt{2}}\right) + C$. This is the final solution to the given integral.

Conclusion

In this article, we evaluated the given integral $\int \cos(2x) \sqrt{1+\sin^2(2x)} \, dx$. We used the given hint to simplify the integral and eventually found its solution. The final solution is $\frac{1}{2}\cos(2x)\sqrt{2-\cos^2(2x)} + \frac{1}{2}\sin^{-1}\left(\frac{\cos(2x)}{\sqrt{2}}\right) + C$.

Final Answer

The final answer is $\boxed{\frac{1}{2}\cos(2x)\sqrt{2-\cos^2(2x)} + \frac{1}{2}\sin^{-1}\left(\frac{\cos(2x)}{\sqrt{2}}\right) + C}$.

Discussion

The given integral appears to be complex and may require some advanced techniques to solve. However, with the given hint, we were able to simplify the integral and eventually find its solution. The final solution is a combination of trigonometric functions and inverse trigonometric functions.

Related Topics

• Trigonometric integrals
• Inverse trigonometric functions
• Advanced calculus techniques

References

• [1] "Calculus" by Michael Spivak
• [2] "Trigonometry" by I.M. Gelfand
• [3] "Advanced Calculus" by Edwin Hewitt

Keywords

• Trigonometric integrals
• Inverse trigonometric functions
• Advanced calculus techniques
• Integral evaluation
• Trigonometry
• Calculus
  

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Introduction

In our previous article, we evaluated the given integral $\int \cos(2x) \sqrt{1+\sin^2(2x)} \, dx$. In this article, we will answer some frequently asked questions related to this integral.

Q: What is the final solution to the given integral?

A: The final solution to the given integral is $\frac{1}{2}\cos(2x)\sqrt{2-\cos^2(2x)} + \frac{1}{2}\sin^{-1}\left(\frac{\cos(2x)}{\sqrt{2}}\right) + C$.

Q: How did you simplify the integral?

A: We used the given hint that $\int \sec^3(x) \, dx = \frac{1}{2}(\sec(x) \tan(x) + \ln|\sec(x) + \tan(x)|) + C$. We also used the trigonometric identity $\sin^2(2x) = 1 - \cos^2(2x)$ to rewrite the integral.

Q: What is the significance of the given hint?

A: The given hint is a well-known result in calculus that can be used to evaluate certain types of integrals. In this case, it helped us simplify the integral and eventually find its solution.

Q: Can you explain the concept of inverse trigonometric functions?

A: Inverse trigonometric functions are functions that return the angle whose trigonometric function is a given value. For example, the inverse sine function returns the angle whose sine is a given value.

Q: How do you evaluate integrals involving inverse trigonometric functions?

A: To evaluate integrals involving inverse trigonometric functions, you can use the following formulas:

• $\int \sin^{-1}(x) \, dx = x\sin^{-1}(x) + \sqrt{1-x^2} + C$
• $\int \cos^{-1}(x) \, dx = x\cos^{-1}(x) - \sqrt{1-x^2} + C$
• $\int \tan^{-1}(x) \, dx = x\tan^{-1}(x) - \frac{1}{2}\ln(1+x^2) + C$

Q: What are some common techniques used to evaluate integrals?

A: Some common techniques used to evaluate integrals include:

• Substitution
• Integration by parts
• Integration by partial fractions
• Trigonometric substitution
• Integration of inverse trigonometric functions

Q: Can you provide some examples of integrals that can be evaluated using these techniques?

A: Yes, here are some examples:

• $\int \sin(x) \cos(x) \, dx = \frac{1}{2}\sin^2(x) + C$
• $\int \frac{1}{x^2+1} \, dx = \tan^{-1}(x) + C$
• $\int \frac{1}{x^2-1} \, dx = \frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| + C$

Conclusion

In this article, we answered some frequently asked questions related to the given integral $\int \cos(2x) \sqrt{1+\sin^2(2x)} \, dx$. We also provided some examples of integrals that can be evaluated using common techniques.

Final Answer

The final answer is $\boxed{\frac{1}{2}\cos(2x)\sqrt{2-\cos^2(2x)} + \frac{1}{2}\sin^{-1}\left(\frac{\cos(2x)}{\sqrt{2}}\right) + C}$.

Discussion

The given integral appears to be complex and may require some advanced techniques to solve. However, with the given hint, we were able to simplify the integral and eventually find its solution. The final solution is a combination of trigonometric functions and inverse trigonometric functions.

Related Topics

• Trigonometric integrals
• Inverse trigonometric functions
• Advanced calculus techniques
• Integral evaluation
• Trigonometry
• Calculus

References

• [1] "Calculus" by Michael Spivak
• [2] "Trigonometry" by I.M. Gelfand
• [3] "Advanced Calculus" by Edwin Hewitt

Keywords

• Trigonometric integrals
• Inverse trigonometric functions
• Advanced calculus techniques
• Integral evaluation
• Trigonometry
• Calculus