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The value of $\int\frac{dx}{\sin^6x}$ is: |
$-\frac{\cot^5x}{5}-2\frac{\cot^3x}{3}-\cot x +C$ $\frac{\cot^5x}{5}+2\frac{\cot^3x}{3}-\cot x +C$ $-\frac{\cot^3x}{5}-2\frac{\cot^5x}{3}-\cot x +C$ None of these |
$-\frac{\cot^5x}{5}-2\frac{\cot^3x}{3}-\cot x +C$ |
$I=\int\frac{dx}{\sin^6x}=\int cosec^4x.cosec^2x\, dx$ [Put cot x - t ⇒ -cosec2x dx = dt] $=\int(\cot^2x+1)^2cosec^2x\,dx=\int(t^2+1)^2(-dt)=-\frac{t^5}{5}-2\frac{t^3}{3}-t+C$ (where t = cot x) |
