## Related questions with answers

For the function in this exercise, find the general antiderivative. Check your answer by differentiation. $f ( x ) = e ^ { \sin x } \cos x$

Solution

Verified$\underline{\bold{Given}}:$

$\begin{align*} f(x) = e^{\mathrm{sin \ \text{$x$}}}\mathrm{cos \ \text{$x$}} \end{align*}$

To do the integral, we first take,

$\begin{align*} u&=\mathrm{sin \ \text{$x$}} & & \text{and} & du&=\mathrm{cos \ \text{$x$}} \ dx \end{align*}$

$\underline{\bold{Solve}}:$

The general anti-derivative of this function is given as

$\begin{align*}\tag{Substitute the values} F(x)&= \int e^{\mathrm{sin \ \text{$x$}}}\mathrm{cos \ \text{$x$}} \ dx\\\\\tag{Integrate} &=\int e^{u} \ du\\\\\tag{Simplify} &=e^{u}+C\\\\\tag{Final result} &=\boxed{e^{\mathrm{sin \ \text{$x$}}}+C} \end{align*}$

$\underline{\bold{Check}}:$

By differentiating the answer which we got, we have

$\begin{align*}\tag{Differentiate using chain rule} \dfrac{d}{dx}\left(e^{\mathrm{sin \ \text{$x$}}}\right)&=\left(e^{\mathrm{sin \ \text{$x$}}} \cdot (\mathrm{cos \ \text{$x$}})\right)\\\\\tag{Hence, proved that LHS=RHS} &=\boxed{e^{\mathrm{sin \ \text{$x$}}} \mathrm{cos \ \text{$x$}}} \ \checkmark \end{align*}$

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