# Evaluate the indefinite integral. (Use C for the constant of

Evaluate the indefinite integral. (Use C for the constant of integration.)
$$\displaystyle\int{e}^{{{\cos{{\left({9}{t}\right)}}}}}{\sin{{\left({9}{t}\right)}}}{\left.{d}{t}\right.}$$

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Howell
Step 1
Given integral:
$$\displaystyle\int{e}^{{{\cos{{\left({9}{t}\right)}}}}}{\sin{{\left({9}{t}\right)}}}{\left.{d}{t}\right.}$$
Step 2
Now,
$$\displaystyle\int{e}^{{{\cos{{\left({9}{t}\right)}}}}}{\sin{{\left({9}{t}\right)}}}{\left.{d}{t}\right.}$$
Sustitute:
$$\displaystyle{\cos{{\left({9}{t}\right)}}}={u}$$
Differentiate both sides:
$$\displaystyle\Rightarrow-{\sin{{\left({9}{t}\right)}}}\times{9}{\left.{d}{t}\right.}={d}{u}\ \ \ {\left[\because{\frac{{{d}{\left({\cos{{\left({x}\right)}}}\right)}}}{{{\left.{d}{x}\right.}}}}=-{\sin{{\left({x}\right)}}}\right]}$$
$$\displaystyle\Rightarrow{\sin{{\left({9}{t}\right)}}}{\left.{d}{t}\right.}=-{\frac{{{d}{u}}}{{{9}}}}$$
Now the integral becomes:
$$\displaystyle\int-{\frac{{{1}}}{{{9}}}}{e}^{{{u}}}{d}{u}=-{\frac{{{1}}}{{{9}}}}\int{e}^{{{u}}}{d}{u}\ \ \ {\left[\because\int{e}^{{{x}}}{\left.{d}{x}\right.}={e}^{{{x}}}+{c}\right]}$$
$$\displaystyle=-{\frac{{{1}}}{{{9}}}}{e}^{{{u}}}+{C}$$
Substitute back the value of u:
$$\displaystyle=-{\frac{{{1}}}{{{9}}}}{e}^{{{\cos{{\left({9}{t}\right)}}}}}+{C}$$
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Roger Noah
Step 1: Use Integration by Substitution.
Let $$\displaystyle{u}={\cos{{9}}}{t},{d}{u}=-{9}{\sin{{9}}}{t}{\left.{d}{t}\right.},\ {t}{h}{e}{n}\ {\sin{{9}}}{t}{\left.{d}{t}\right.}=-{\frac{{{1}}}{{{9}}}}{d}{u}$$
Step 2: Using u and du above, rewrite $$\displaystyle\int{e}^{{{\cos{{9}}}{t}}}{\sin{{9}}}{t}{\left.{d}{t}\right.}$$.
$$\displaystyle\int-{\frac{{{e}^{{{u}}}}}{{{9}}}}{d}{u}$$
Step 3: Use Constant Factor Rule: $$\displaystyle\int{c}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={c}\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$$.
$$\displaystyle-{\frac{{{1}}}{{{9}}}}\int{e}^{{{u}}}{d}{u}$$
Step 4: The integral of $$\displaystyle{e}^{{{x}}}\ {i}{s}\ {e}^{{{x}}}$$.
$$\displaystyle-{\frac{{{e}^{{{u}}}}}{{{9}}}}$$
Step 5: Substitute $$\displaystyle{u}={\cos{{9}}}{t}$$ back into the original integral.
$$\displaystyle-{\frac{{{e}^{{{\cos{{9}}}{t}}}}}{{{9}}}}$$
$$\displaystyle-{\frac{{{e}^{{{\cos{{9}}}{t}}}}}{{{9}}}}+{C}$$