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Question

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

Solution

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Answered 4 months ago
Answered 4 months ago
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Given:\underline{\bold{Given}}:

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

To do the integral, we first take,

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

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

The general anti-derivative of this function is given as

F(x)=esin xcos x dx=eu du=eu+C=esin x+C\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*}

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

By differentiating the answer which we got, we have

ddx(esin x)=(esin x(cos x))=esin xcos x \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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