Question Statement

Evaluate the indefinite integral:

$\int 9\tan (x+7)dx$

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Background and Explanation

This problem involves integrating a tangent function with a linear argument using u-substitution (change of variables). You will also need the standard integral formula for $\tan x$, which results in a logarithmic function involving $\sec x$.

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Solution

We begin by factoring out the constant 9 using the constant multiple rule for integration:

$I=\int 9\tan (x+7)dx=9\int \tan (x+7)dx$

To integrate $\tan (x+7)$, we use substitution to simplify the argument of the tangent function. Put:

$t=x+7$

Taking differentials of both sides, we get $d(x+7)=dt$, which simplifies to:

$dx+0=dt\Rightarrow dx=dt$

Now we substitute $t$ and $dt$ into our integral:

$I=9\int \tan tdt$

We apply the standard integral formula $\int \tan xdx=\ln (\sec x)+C$ (where $C$ is the constant of integration):

$I=9\ln (\sec t)+c$

Finally, we substitute back $t=x+7$ to express the answer in terms of the original variable $x$:

$I=9\ln (\sec (x+7))+c$

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Key Formulas or Methods Used

• Constant Multiple Rule: $\int k\cdot f(x)dx=k\int f(x)dx$
• Substitution Method: If $u=g(x)$, then $du=g^{\prime }(x)dx$
• Standard Integral: $\int \tan xdx=\ln (\sec x)+C$ (or equivalently $-\ln (\cos x)+C$)

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Summary of Steps

1. Factor out the constant 9 from the integral.
2. Substitute $t=x+7$, which gives $dt=dx$.
3. Rewrite the integral as $9\int \tan tdt$.
4. Apply the standard integral formula to get $9\ln (\sec t)+c$.
5. Substitute back $t=x+7$ to obtain the final answer: $9\ln (\sec (x+7))+c$.