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$\int\frac{\sin^2x}{\cos^6x}dx$ is a: |
Polynomial of degree 5 in sin x Polynomial of degree 4 in tan x Polynomial of degree 5 in tan x Polynomial of degree 5 in cos x |
Polynomial of degree 5 in tan x |
$I=\int\frac{\sin^2x}{\cos^6x}dx=\int\tan^2x\sec^2x.\sec^2x\,dx=\int\tan^2x(1+\tan^2x).\sec^2 x\,dx$ Put tan x = t to get: $I=\int t^2(t^2+1)dt=\frac{t^5}{5}+\frac{t^3}{3}+C=\frac{\tan^5x}{5}+\frac{\tan^3x}{3}+C$ |
