Question

If $\int \frac{1-(\cot x{)}^{2019}}{\tan x+(\cot x{)}^{2020}}dx=\frac{1}{n}\ln \left|(f(x){)}^n+(g(x){)}^n\right|+c$, then the value of $n{\left[(f(x){)}^4+(g(x){)}^4\right]}_{x=\frac{\pi }{3}}=$

• A. $\frac{10105}{16}$
• B. $\frac{10012}{15}$
• C. $\frac{20210}{9}$
• D. $\frac{10105}{8}$

Answer 1

Answer: (D)

Let $I=\int \frac{1-(\cot x{)}^{2019}}{\tan x+(\cot x{)}^{2020}}dx$
$=\int \frac{1-\frac{(\cos x{)}^{2019}}{(\sin x{)}^{2019}}}{\frac{\sin x}{\cos x}+\frac{(\cos x{)}^{2020}}{(\sin x{)}^{2020}}dx}$
$=\int \frac{{\sin }^{2019}x-{\cos }^{2019}x}{{\sin }^{2021}x+{\cos }^{2021}x}\cdot \sin x\cos xdx$
Put
${\sin }^{2021}x+{\cos }^{2021}x=t$
$\Rightarrow \left[2021{\sin }^{2020}x\cdot \cos x+2021{\cos }^{2020}\cdot (-\sin x)\right]$
$dx=dt$
$\Rightarrow \sin x\cos x\left[{\sin }^{2019}x-{\cos }^{2019}x\right]dx=\frac{1}{2021}dt$
$\therefore I=\frac{1}{2021}\int \frac{1}{t}dt=\frac{1}{2021}\log t+C$
$=\frac{1}{2021}\log \left[{\sin }^{2021}x+{\cos }^{2021}x\right]+C$
$\therefore f(x)=\sin x,g(x)=\cos x,n=2021$
$\therefore n[{\left(f(x{)}^4+\left(g(x{)}^4\right)\right]}_{x=\frac{\pi }{3}}=2021\left[{\sin }^4\frac{\pi }{3}+{\cos }^4\frac{\pi }{3}\right]$
$=2021\left[{\left(\frac{\sqrt{3}}{2}\right)}^4+{\left(\frac{1}{2}\right)}^4\right]$
$=2021\left[\frac{9}{16}+\frac{1}{16}\right]=\frac{2021\times 10}{16}=\frac{10105}{8}$