Integration Mania

$\displaystyle \int _{ \dfrac { \pi }{ 4 } }^{ \dfrac { \pi }{ 3 } } e^x(\dfrac{2}{2cos^2x} + \dfrac{2sinxcos}{2cos^2x})$

$\displaystyle \int _{ \dfrac { \pi }{ 4 } }^{ \dfrac { \pi }{ 3 } } e^x(sec^2x + tanx)$

$\displaystyle \int e^x(f(x) + f'(x))dx = e^xf(x) + C$

$= \big[_{\frac{ \pi }{ 4 } }^{ \frac { \pi }{ 3 } } e^xtanx \big]$

$= e^{\dfrac{\pi}{3}}\sqrt{3} - e^{\dfrac{\pi}{4}}$

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Note-

$\displaystyle \int e^x(f(x) + f'(x))dx = e^xf(x) + C$

Proof - By parts , take $e^x$ as first function

$\displaystyle \int e^x f(x)dx = e^xf(x) + C - \int e^x f'(x)dx$

$\boxed{\displaystyle \int e^x(f(x) + f'(x))dx = e^xf(x) + C}$

$\begin{aligned} I & = \int_\frac \pi 4 ^\frac \pi 3 e^x \left(\frac {2+\sin(2x)}{1+\cos(2x)} \right) dx \\ & = \int_\frac \pi 4 ^\frac \pi 3 e^x \left(\frac {2+2\sin x \cos x}{1+2 \cos^2 x-1} \right) dx \\ & = \int_\frac \pi 4 ^\frac \pi 3 e^x \left(\frac {2+2\sin x \cos x}{2\cos^2 x} \right) dx \\ & = \int_\frac \pi 4 ^\frac \pi 3 e^x \left(\sec^2 x + \tan x \right) dx \\ & = \color{#3D99F6}{\int_\frac \pi 4 ^\frac \pi 3 e^x \sec^2 x \ dx} + \int_\frac \pi 4 ^\frac \pi 3 e^x \tan x \ dx & \small \color{#3D99F6}{\text{By integration by parts,}} \\ & = \color{#3D99F6}{e^x \tan x \bigg|_\frac \pi 4 ^\frac \pi 3 - \cancel{\int_\frac \pi 4 ^\frac \pi 3 e^x \tan x \ dx}} + \cancel{\int_\frac \pi 4 ^\frac \pi 3 e^x \tan x \ dx} & \small \color{#3D99F6}{f(x) = e^x, \ g'(x) = \sec^2 x} \\ & = e^\frac \pi 3 \tan \frac \pi 3 - e^\frac \pi 4 \tan \frac \pi 4 \\ & = \sqrt 3 e^\frac \pi 3 - e^\frac \pi 4 \\ & = e^\frac \pi 4 \left(\sqrt 3 e^\frac \pi{12} - 1\right) \end{aligned}$

$\implies \dfrac {b^2c}{ad^{10}} = \dfrac {(\sqrt 3)^2 \cdot 12}{4 \cdot 1^{10}} = \boxed{9}$