Question

If $\int \frac{\sin x+\cos x}{9+16\sin 2x}dx=A\ln \left|\frac{4\sin x-4\cos x+B}{4\cos x-4\sin x+P}\right|+C,$ then which of the following is/are true?
(where $A,B,P$ are fixed constants and $C$ is integration constant)

• A. $BP=25$
• B. $AP=\frac{1}{9}$
• C. $AB=\frac{1}{8}$
• D. $B+P=\frac{1}{4A}$

Answer 1

Answer: (A), (C), (D)

$B+P=\frac{1}{4A}$

$\int \frac{\sin x+\cos x}{9+16\sin 2x}dx$
Put $\sin x-\cos x=t$
$\Rightarrow (\cos x+\sin x)dx=dt$
Also, $\sin 2x=1-t^2$
$\therefore I=\int \frac{1}{9+16(1-t^2)}dt=\int \frac{1}{25-16t^2}dt=\int \frac{1}{(5{)}^2-(4t{)}^2}dt=\frac{1}{4}\cdot \frac{1}{10}\ln \left|\frac{5+4t}{5-4t}\right|+C[\because \int \frac{1}{a^2-x^2}dx=\frac{1}{2a}\ln \left|\frac{a+x}{a-x}\right|+C]=\frac{1}{40}\ln \left|\frac{5+4(\sin x-\cos x)}{5-4(\sin x-\cos x)}\right|+C\Rightarrow A=\frac{1}{40},B=5,P=5$