Biết rằng tích phân $\large I=\int _0^{\dfrac{\pi}{4}}\dfrac{\ln (\sin

Biết rằng tích phân $\large I=\int _0^{\dfrac{\pi}{4}}\dfrac{\ln (\sin x+15\cos x)}{\cos^2x}dx=a\pi +b\ln 2+c\ln 3+d\ln 5$, trong đó $\large a,b,c,d\in \mathbb{Q}$. Tính $\large T=a+b+c+d.$

• A.

  $\large T=\dfrac{133}{4}$

• B.

  $\large T=\dfrac{313}{4}$

• C.

  $\large T=\dfrac{135}{4}$

• D.

  $\large T=\dfrac{195}{4}$

Chọn A

Ta có: $\large \left\{\begin{align}& u=\ln (\sin x+15\cos x)\\& dv=\dfrac{1}{\cos^2x}dx\\\end{align}\right. $ $\large \Rightarrow \left\{\begin{align}& du=\dfrac{\cos x-15\sin x}{\sin x+15\cos x}\\& v=\tan x+15=\dfrac{\sin x+15\cos x}{\cos x}\\\end{align}\right.$

Suy ra: 

$\large I=\left. (\tan x+15)\ln (\sin x+15\cos x)\right|_0^{\dfrac{\pi}{4}}-\int_0^{\dfrac{\pi}{4}}\dfrac{\cos x-15\sin x}{\cos x}$

$\large =16\ln 8\sqrt{2}-15\ln 15-\int_0^{\dfrac{\pi}{4}} dx+15\int_0^{\dfrac{\pi}{4}}\dfrac{\sin x}{\cos x}dx=16\ln 8\sqrt{2}-15\ln 15-\dfrac{\pi}{4}-15\int_0^{\dfrac{\pi}{4}}\dfrac{d(\cos x)}{\cos x}$

$\large =16\ln 8\sqrt{2}-15\ln 15-\dfrac{\pi}{4}-\left. 15\ln |\cos x|\right|^{\dfrac{\pi}{4}}_0=16\ln 8\sqrt{2}-15\ln 15-\dfrac{\pi}{4}-15\ln \dfrac{1}{\sqrt{2}}$

$\large =-\dfrac{1}{4}\pi+\dfrac{127}{2}\ln 2-15\ln 3-15\ln 5$

Vậy $\large T=a+b+c+d=-\dfrac{1}{4}+\dfrac{127}{2}-15-15=\dfrac{133}{4}$