Question

Solve the definite integral.

${\int }_{-1}^1{2}^x\cos xdx$

Short answer

The solution is$\frac{\ln \left(2\right)}{\ln \left(2\right)+1}\left[\left\{\frac{2}{\ln \left(2\right)}\left(\cos \left(1\right)+\frac{\sin \left(1\right)}{\ln \left(2\right)}\right)\right\}-\frac{1}{2\ln \left(2\right)}\left\{\cos (-1)+\frac{\sin (-1)}{\ln \left(2\right)}\right\}\right]$.

Step-by-step solution

Step 1. Given information.

The given integral is${\int }_{-1}^1{2}^x\cos xdx$.

Step 2. First, solve the indefinite integral.

$\begin{array}{rcl}I& =& \int 2^x\cos xdx\\ & =& \cos x\int 2^{x}dx-\int \left[\frac{d}{dx}\left(\cos x\right)\int 2^{x}dx\right]dx\\ & =& \cos x·\frac{2^x}{\ln \left(2\right)}-\int -\sin x·\frac{2^x}{\ln \left(2\right)}dx\\ & =& \cos x·\frac{2^x}{\ln \left(2\right)}+\left[\frac{\sin x}{\ln \left(2\right)}\int 2^{x}dx-\int \left\{\frac{d}{dx}\left(\sin x\right)\int 2^{x}dx\right\}dx\right]\\ & =& \cos x·\frac{2^x}{\ln \left(2\right)}+\frac{\sin x}{\ln \left(2\right)}·\frac{2^x}{\ln \left(2\right)}-\int \cos x·\frac{2^x}{\ln \left(2\right)}dx\\ & =& \cos x·\frac{2^x}{\ln \left(2\right)}+\frac{\sin x}{\ln \left(2\right)}·\frac{2^x}{\ln \left(2\right)}-\frac{1}{\ln 2}\int 2^x\cos xdx\end{array}$

Step 3. Continued the above solution.

$\begin{array}{rcl}I& =& \cos x·\frac{2^x}{\ln \left(2\right)}+\frac{\sin x}{\ln \left(2\right)}·\frac{2^x}{\ln \left(2\right)}-\frac{1}{\ln 2}\int 2^x\cos xdx\\ I& =& \cos x·\frac{2^x}{\ln \left(2\right)}+\frac{\sin x}{\ln \left(2\right)}·\frac{2^x}{\ln \left(2\right)}-\frac{1}{\ln 2}I\\ I+\frac{1}{\ln \left(2\right)}I& =& \cos x·\frac{2^x}{\ln \left(2\right)}+\frac{\sin x}{\ln \left(2\right)}·\frac{2^x}{\ln \left(2\right)}\\ I\left(1+\frac{1}{\ln \left(2\right)}\right)& =& \cos x·\frac{2^x}{\ln \left(2\right)}+\frac{\sin x}{\ln \left(2\right)}·\frac{2^x}{\ln \left(2\right)}\\ I\left(\frac{\ln \left(2\right)+1}{\ln \left(2\right)}\right)& =& \cos x·\frac{2^x}{\ln \left(2\right)}+\frac{\sin x}{\ln \left(2\right)}·\frac{2^x}{\ln \left(2\right)}\\ I& =& \frac{\ln \left(2\right)}{\ln \left(2\right)+1}\left[\cos x·\frac{2^x}{\ln \left(2\right)}+\frac{\sin x}{\ln \left(2\right)}·\frac{2^x}{\ln \left(2\right)}\right]\end{array}$

Step 4. Now apply the limit of integration.

$\begin{array}{rcl}{\int }_{-1}^1{2}^x\cos xdx& =& \frac{\ln \left(2\right)}{\ln \left(2\right)+1}{\left[\cos x·\frac{2^x}{\ln \left(2\right)}+\frac{\sin x}{\ln \left(2\right)}·\frac{2^x}{\ln \left(2\right)}\right]}_{-1}^1\\ & =& \frac{\ln \left(2\right)}{\ln \left(2\right)+1}\left[\left\{\cos \left(1\right)·\frac{2^1}{\ln \left(2\right)}+\frac{\sin \left(1\right)}{\ln \left(2\right)}·\frac{2^1}{\ln \left(2\right)}\right\}-\left\{\cos (-1)·\frac{2^{-1}}{\ln \left(2\right)}+\frac{\sin (-1)}{\ln \left(2\right)}·\frac{2^{(-1)}}{\ln \left(2\right)}\right\}\right]\\ & =& \frac{\ln \left(2\right)}{\ln \left(2\right)+1}\left[\left\{\frac{2}{\ln \left(2\right)}\left(\cos \left(1\right)+\frac{\sin \left(1\right)}{\ln \left(2\right)}\right)\right\}-\left\{\cos (-1)·\frac{1}{2\ln \left(2\right)}+\frac{\sin (-1)}{\ln \left(2\right)}·\frac{1}{2\ln \left(2\right)}\right\}\right]\\ & =& \frac{\ln \left(2\right)}{\ln \left(2\right)+1}\left[\left\{\frac{2}{\ln \left(2\right)}\left(\cos \left(1\right)+\frac{\sin \left(1\right)}{\ln \left(2\right)}\right)\right\}-\frac{1}{2\ln \left(2\right)}\left\{\cos (-1)+\frac{\sin (-1)}{\ln \left(2\right)}\right\}\right]\end{array}$

Step 5. Simplified answer.

Hence, the required value is $\frac{\ln \left(2\right)}{\ln \left(2\right)+1}\left[\left\{\frac{2}{\ln \left(2\right)}\left(\cos \left(1\right)+\frac{\sin \left(1\right)}{\ln \left(2\right)}\right)\right\}-\frac{1}{2\ln \left(2\right)}\left\{\cos (-1)+\frac{\sin (-1)}{\ln \left(2\right)}\right\}\right]$.