# How do you find the indefinite integral of int 2^(sinx)cosx?

Dec 24, 2016

$\int {2}^{\sin x} \cos x \text{ } \mathrm{dx} = {2}^{\sin} \frac{x}{\ln} 2 + C$.

#### Explanation:

Let $u = \sin x$.
Then $\frac{\mathrm{du}}{\mathrm{dx}} = \cos x$, and thus $\mathrm{du} = \cos x \text{ } \mathrm{dx}$.

Substituting this into the given integral, we get

$\int {2}^{\sin x} \cos x \text{ "dx=int 2^u" } \mathrm{du}$
color(white)(int 2^(sin x) cos x " "dx)=color(navy)(1/(ln 2))int color(navy)(ln 2) * 2^u" "du
$\textcolor{w h i t e}{\int {2}^{\sin x} \cos x \text{ } \mathrm{dx}} = \frac{1}{\ln 2} \cdot {2}^{u} + C$

And since $u = \sin x$, we substitute back:

$\textcolor{w h i t e}{\int {2}^{\sin x} \cos x \text{ } \mathrm{dx}} = \frac{1}{\ln 2} \cdot {2}^{\sin} x + C$
$\textcolor{w h i t e}{\int {2}^{\sin x} \cos x \text{ } \mathrm{dx}} = {2}^{\sin} \frac{x}{\ln 2} + C$

So $\int {2}^{\sin x} \cos x \text{ } \mathrm{dx} = {2}^{\sin} \frac{x}{\ln} 2 + C$.

## Check:

Using the chain rule and the exponential rule for derivatives:
$\frac{d}{\mathrm{dx}} \left({a}^{u}\right) = \ln a \cdot {a}^{u} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

We get

$\frac{d}{\mathrm{dx}} \left({2}^{\sin x} / \ln 2 + C\right) = \frac{1}{\ln} 2 \cdot \ln 2 \cdot {2}^{\sin} x \cdot \cos x$
$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} \left({2}^{\sin x} / \ln 2\right)} = {2}^{\sin} x \cdot \cos x$,

which matches our integrand from above.