Question

$I:\int \frac{dx}{a^2{\sin }^{2}x+b^2{\cos }^{2}x}=\frac{1}{ab}ta{n}^{-1}\left(\frac{a\tan x}{b}\right)+c$

$II:\int \frac{3cosx+2\sin x}{4\sin x+5\infty sx}dx=\frac{23x}{41}+\frac{2}{41}ln(|4\sin x+5cosx|)+c$

• A. only I is true
• B. only II is true
• C. both I and II are true
• D. neither I nor II are true

Answer 1

The correct option is C both I and II are true

$I=\int \frac{dx}{a^2{\sin }^{2}x+b^2{\cos }^{2}x}$ $=\int \frac{dx({\sec }^{2}x\times \frac{a}{b})\times \frac{b}{a}}{(1+\frac{a^2}{b^2}{\tan }^{2}x)b^2}$

$z=\frac{a}{b}\tan x$

$dz=\frac{a}{b}{\sec }^{2}xdx$

$I=\int \frac{\frac{b}{a}dz}{{1+z}^2}\times \frac{1}{b^2}$

$I=\frac{1}{ab}\int \frac{dz}{{1+z}^2}$

$I=\frac{1}{ab}{\tan }^{-1}\left(\frac{a\tan x}{b}\right)$

$I^{\prime }=\int \frac{3\cos x+2\sin x}{4\sin x+5\cos x}dx$

$3\cos x+2\sin x=a\left(4\sin x+5\cos x\right)+b\left(4\sin x-5\cos x\right)$

$3=5a+4b$

$2=4a+5b$

On solving the above equations, we have

$b=\frac{2}{41},a=\frac{23}{41}$

$I^{\prime }=\int a+b\frac{f^{\prime }\left(x\right)}{f\left(x\right)}$

$=ax+b\ln |f\left(x\right)|+c$

$=ax+b\ln |4\sin x+5\cos x|+c$

So both are correct.