Question

$\int \frac{\sin x\cos x}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}dx$

• A. $\frac{1}{(b^2-a^2)}\log (a^2{\cos }^{2}x+b^2{\sin }^{2}x).$
• B. $\frac{1}{2(a^2-b^2)}\log (a^2{\cos }^{2}x+b^2{\sin }^{2}x).$
• C. $\frac{1}{2(b^2-a^2)}\log (a^2{\cos }^{2}x+b^2{\sin }^{2}x).$
• D. $\frac{1}{2(b^2-a^2)}\log (a^2{\cos }^{2}x-b^2{\sin }^{2}x).$

Answer 1

Answer: (C)

$\frac{1}{2(b^2-a^2)}\log (a^2{\cos }^{2}x+b^2{\sin }^{2}x).$

Let $I=\int \frac{\sin x\cos x}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}dx$
Put $a^2{\cos }^{2}x+b^2{\sin }^{2}x=t\Rightarrow 2(b^2-a^2)\sin x\cos xdx=dt$
Therefore
$I=\frac{1}{2(b^2-a^2)}\int \frac{dt}{t}=\frac{1}{2(b^2-a^2)}\log t=\frac{1}{2(b^2-a^2)}\log (a^2{\cos }^{2}x+b^2{\sin }^{2}x)$