CUET Preparation Today
If $f(x)=\frac{1}{4-3\cos^2x+5\sin^2x}$ and its anti-derivative $F(x)=\frac{1}{3}\tan^{-1}(g(x))+C$ then g(x) = |
$3 \tan x$ $\sqrt{2}\tan x$ $2 \tan x$ None of these |
$3 \tan x$ |
$F(x)=\frac{1}{4-3\cos^2x+5\sin^2x}=\int\frac{\sec^2x\,dx}{4(1+\tan^2x)-3+5\tan^2x}$ Put tan x = t to get $F(x)=\int\frac{dt}{9t^2+1}+C=\frac{1}{3}\tan^{-1}(3t)+C=\frac{1}{3}\tan^{-1}(3\tan x)+C$ |
