Question

$\int \frac{\mathrm{dx}}{{\sin }^{2}x-12{\cos }^{2}x+\mathrm{cosx}\mathrm{sinx}}$ is equal to

• A. $\frac{1}{7}\ln \left|\frac{\tan x+3}{\tan x-4}\right|+C$
• B. $\frac{1}{7}\ln \left|\frac{\tan x-3}{\tan x+4}\right|+C$
• C. $\frac{1}{7}\ln \left|\frac{\tan x+3}{\tan x+4}\right|+C$
• D. $\frac{1}{7}\ln \left|\frac{\tan x-3}{\tan x-4}\right|+C$

Answer 1

The correct option is B $\frac{1}{7}\ln \left|\frac{\tan x-3}{\tan x+4}\right|+C$

Solving the intregral of the form $I=\int \frac{\mathrm{dx}}{{\mathrm{acos}}^{2}x+{\mathrm{bsin}}^{2}x+\mathrm{cosx}\mathrm{sinx}}$
requires multiplying both numerator and denominator of the integrand by ${\sec }^{2}x$ and substituting $t=\mathrm{tanx}$
Here, multiplying the numerator and denominator of the integrand by ${\sec }^{2}x$ we get,
$I=\int \frac{{\sec }^{2}x\mathrm{dx}}{{\sin }^{2}x{\sec }^{2}x+\mathrm{cosx}\mathrm{sinx}{\sec }^{2}x-12{\cos }^{2}x{\sec }^{2}x}$ $\Rightarrow I=\int \frac{{\sec }^{2}x\mathrm{dx}}{{\tan }^{2}x+\mathrm{tanx}-12}$
Now, substituting $t=\mathrm{tanx}$
we get, $\mathrm{dt}={\sec }^{2}x\mathrm{dx}$
Substituting these back in the integral we get,
$I=\int \frac{\mathrm{dt}}{t^2+t-12}\Rightarrow I=\int \frac{\mathrm{dt}}{(t-3)(t+4)}\Rightarrow I=\frac{1}{7}\int \left(\frac{1}{t-3}-\frac{1}{t+4}\right]\mathrm{dt}\Rightarrow I=\frac{1}{7}\left(\ln \left(t-3\right|-\ln \left(t+4\right|\right]+C\Rightarrow I=\frac{1}{7}\ln \left(\frac{t-3}{t+4}\right|+C$
Now, substituting back $t=\mathrm{tanx}$
we get $I=\frac{1}{7}\ln \left(\frac{\mathrm{tanx}-3}{\mathrm{tanx}+4}\right|+C$
Thus, Option b. is the correct.