Đáp án:
$T= \dfrac{133}{4}$
Giải thích các bước giải:
$\quad I = \displaystyle\int\limits_0^{\tfrac{\pi}{4}}\dfrac{\ln(\sin x +15\cos x)}{\cos^2x}dx$
Đặt $\begin{cases}u = \ln(\sin x +15\cos x)\\dv = \dfrac{1}{\cos^2x}dx\end{cases}\longrightarrow \begin{cases}du =\dfrac{\cos x - 15\sin x}{\sin x +15\cos x}dx\\v = \tan x\end{cases}$
Ta được:
$\quad I = \tan x.\ln(\sin x +15\cos x)\Bigg|_0^{\tfrac{\pi}{4}} - \displaystyle\int\limits_0^{\tfrac{\pi}{4}}\tan x\cdot\dfrac{\cos x - 15\sin x}{\sin x +15\cos x}dx$
$\to I = \ln\left(\sin\dfrac{\pi}{4} + 15\cos\dfrac{\pi}{4}\right) - \displaystyle\int\limits_0^{\tfrac{\pi}{4}}\tan x\cdot\dfrac{1 - 15\tan x}{\tan x +15}dx$
$\to I = \ln8\sqrt2 - \displaystyle\int\limits_0^{\tfrac{\pi}{4}}\left(226 - 15\tan x - \dfrac{3390}{\tan x +15}\right)dx$
$\to I = \ln8\sqrt2 - \left(15\ln\cos x + 226x\right)\Bigg|_0^{\tfrac{\pi}{4}} + 3390\displaystyle\int\limits_0^{\tfrac{\pi}{4}}\dfrac{1}{\tan x +15}dx$
$\to I = \ln8\sqrt2 - 15\ln\dfrac{\sqrt2}{2} - \dfrac{113\pi}{2} + 3390\displaystyle\int\limits_0^{\tfrac{\pi}{4}}\dfrac{\cos x}{\sin x +15\cos x}dx$
$\to I = 11\ln2 - \dfrac{113\pi}{2} + 15\displaystyle\int\limits_0^{\tfrac{\pi}{4}}\dfrac{\cos x - 15\sin x + 15(\sin x + 15\cos x)}{\sin x + 15\cos x}dx$
$\to I = 11\ln2 - \dfrac{113\pi}{2} + 15\left[\ln(\sin x + 15\cos x) + 15x\right]\Bigg|_0^{\tfrac{\pi}{4}}$
$\to I = 11\ln2 - \dfrac{113\pi}{2} + 15\left(\ln8\sqrt2 - \ln15 + \dfrac{15\pi}{4}\right)$
$\to I = -\dfrac{\pi}{4} + \dfrac{127}{2}\ln2 - 15\ln3 - 15\ln5$
$\to \begin{cases}a = -\dfrac14\\b = \dfrac{127}{2}\\c = -15\\d = -15\end{cases}$
$\to a + b + c + d = -\dfrac14 +\dfrac{127}{2} - 15 - 15$
$\to T= \dfrac{133}{4}$
