How do you integrate $\int \frac{e^{x}}{\sqrt{e^{2 x} + 16}} d x$ $int (e^x)/sqrt(e^(2x) +16)dx$ using trigonometric substitution?

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Answer 1 · 2018-03-14T06:05:13

$\int \frac{e^{x}}{\sqrt{e^{2 x} + 16}} d x = ln ∣ ∣ \sqrt{e^{2 x} + 16} + e^{x} ∣ ∣ + c$ $inte^x/sqrt(e^(2x)+16)dx=ln|sqrt(e^(2x)+16)+e^x|+c$

Let $e^{x} = 4 tan u$ $e^x=4tanu$ then $e^{x} d x = 4 {sec}^{2} u d u$ $e^xdx=4sec^2udu$ and

$d x = \frac{4 {sec}^{2} u}{4 tan u} d u$ $dx=(4sec^2u)/(4tanu)du$ -as $e^{x} = 4 tan u$ $e^x=4tanu$

and $\sqrt{e^{2 x} + 16} = \sqrt{16 {tan}^{2} u + 16} = 4 sec u$ $sqrt(e^(2x)+16)=sqrt(16tan^2u+16)=4secu$

Hence $\int \frac{e^{x}}{\sqrt{e^{2 x} + 16}} d x$ $inte^x/sqrt(e^(2x)+16)dx$

= $\int \frac{4 tan u}{4 sec u} \frac{4 {sec}^{2} u}{4 tan u} d u$ $int(4tanu)/(4secu)(4sec^2u)/(4tanu)du$

= $\int sec u d u$ $intsecudu$

= $ln | sec u + tan u | + c_{1}$ $ln|secu+tanu|+c_1$

= $ln ∣ ∣ \sqrt{{tan}^{2} u + 1} + tan u ∣ ∣ + c_{1}$ $ln|sqrt(tan^2u+1)+tanu|+c_1$

= $ln ∣ ∣ ∣ \frac{\sqrt{e^{2 x} + 16}}{4} + \frac{e^{x}}{4} ∣ ∣ ∣ + c_{1}$ $ln|sqrt(e^(2x)+16)/4+e^x/4|+c_1$

= $ln ∣ ∣ \sqrt{e^{2 x} + 16} + e^{x} ∣ ∣ + c$ $ln|sqrt(e^(2x)+16)+e^x|+c$ , where $c$ $c$ is another constant.

How do you integrate int (e^x)/sqrt(e^(2x) +16)dx∫ex√e2x+16dxint (e^x)/sqrt(e^(2x) +16)dx using trigonometric substitution?