Question Statement

Evaluate the integral: $\int (\tan 5x+\cos 7x)dx$

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Background and Explanation

This problem requires integrating basic trigonometric functions using standard integration formulas and the substitution method (reverse chain rule). You need to know how to integrate $\tan (ax)$ and $\cos (ax)$, either by memorizing the standard results or deriving them via substitution.

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Solution

We solve this by splitting the integral into two separate integrals. There are two common approaches for handling the $\tan 5x$ term—one using explicit substitution and another using the standard integral formula.

Method 1: Substitution Approach

Begin by separating the integral using the sum rule: $I=\int (\tan 5x+\cos 7x)dx=\int \tan 5xdx+\int \cos 7xdx$

For the cosine term: Using the standard integral $\int \cos (ax)dx=\frac{\sin (ax)}{a}$: $\int \cos 7xdx=\frac{\sin 7x}{7}$

For the tangent term: Rewrite $\tan 5x$ as $\frac{\sin 5x}{\cos 5x}$: $\int \tan 5xdx=\int \frac{\sin 5x}{\cos 5x}dx$

Notice that the derivative of $\cos 5x$ is $-5\sin 5x$. To match the form $\int \frac{f^{\prime }(x)}{f(x)}dx=\ln (f(x))$, we multiply and divide by $-5$: $=\frac{-1}{5}\int \frac{-5\sin 5x}{\cos 5x}dx$

Applying the logarithmic integration rule: $=\frac{-1}{5}\ln (\cos 5x)$

Combining both results: $I=\frac{-1}{5}\ln (\cos 5x)+\frac{1}{7}\sin 7x+c$

We can simplify using logarithm properties. Since $-\ln (\cos 5x)=\ln ((\cos 5x{)}^{-1})=\ln \left(\frac{1}{\cos 5x}\right)=\ln (\sec 5x)$: $I=\frac{1}{5}\ln (\sec 5x)+\frac{1}{7}\sin 7x+c$

Method 2: Direct Formula Application

Alternatively, apply the standard integral $\int \tan xdx=\ln (\sec x)$ directly with the chain rule adjustment:

$I=\int \tan 5xdx+\int \cos 7xdx$

Using $\int \tan (ax)dx=\frac{\ln (\sec (ax))}{a}$ and $\int \cos (ax)dx=\frac{\sin (ax)}{a}$: $=\frac{\ln (\sec 5x)}{5}+\frac{\sin 7x}{7}+c$

$=\frac{1}{5}\ln (\sec 5x)+\frac{1}{7}\sin 7x+c$

Verification by Differentiation

To verify, differentiate the result $\frac{1}{5}\ln (\sec 5x)+\frac{1}{7}\sin 7x+c$:

$=\frac{1}{5}\cdot \frac{d}{dx}(\ln (\sec 5x))+\frac{1}{7}\cdot \frac{d}{dx}(\sin 7x)+\frac{d}{dx}(c)$

Apply the chain rule to each term: $=\frac{1}{5}\cdot \frac{1}{\sec 5x}\cdot \frac{d}{dx}(\sec 5x)+\frac{1}{7}\cdot \cos 7x\cdot 7+0$

Compute the derivatives (remembering $\frac{d}{dx}(\sec u)=\sec u\tan u\cdot \frac{du}{dx}$): $=\frac{1}{5}\cdot \frac{1}{\sec 5x}\cdot (\sec 5x\tan 5x)\cdot 5+\frac{1}{7}\cdot \cos 7x\cdot 7$

Simplify by canceling the 5s in the first term and the 7s in the second: $=\tan 5x+\cos 7x$

This matches the original integrand, confirming the solution is correct.

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Key Formulas or Methods Used

• Sum Rule for Integration: $\int [f(x)+g(x)]dx=\int f(x)dx+\int g(x)dx$
• Standard Integral: $\int \cos (ax)dx=\frac{\sin (ax)}{a}+C$
• Standard Integral: $\int \tan (ax)dx=\frac{\ln |\sec (ax)|}{a}+C=-\frac{\ln |\cos (ax)|}{a}+C$
• Logarithmic Integration: $\int \frac{f^{\prime }(x)}{f(x)}dx=\ln |f(x)|+C$
• Logarithm Properties: $-\ln (u)=\ln (u^{-1})=\ln \left(\frac{1}{u}\right)$

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Summary of Steps

1. Separate the integral into $\int \tan 5xdx+\int \cos 7xdx$
2. Integrate $\cos 7x$ using the standard formula to obtain $\frac{\sin 7x}{7}$
3. Integrate $\tan 5x$ by either:
   • Rewriting as $\frac{\sin 5x}{\cos 5x}$ and using substitution (multiplying by $\frac{-5}{-5}$), yielding $-\frac{1}{5}\ln (\cos 5x)$, or
   • Using the direct formula $\frac{\ln (\sec 5x)}{5}$
4. Combine and simplify using logarithm properties to get $\frac{1}{5}\ln (\sec 5x)+\frac{1}{7}\sin 7x+c$
5. Verify by differentiating the result to recover $\tan 5x+\cos 7x$