If ∫$$cot$$⁡xsin⁡xcos⁡$$xdx=Pcot$$⁡$$x+Q$$ then the value of $$P4$$ is

Let $$I=$$∫$$cot$$⁡xsin⁡xcos⁡$$xdx=$$∫$$cot$$⁡xtan⁡$$xsec2$$⁡$$xdx=$$∫$$sec2$$⁡$$x($$$$tan$$⁡$$x)3/2dx=$$−$$2tan$$⁡$$x+Q=$$−$$2cot$$⁡$$x+Q$$∴ $$P=$$−$$2$$⇒$$P4=16$$

Methods of integration

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Question

If $\int \frac{\sqrt{\cot x}}{\sin xcos x} dx = P \sqrt{\cot x} + Q$ then the value of P⁴ is

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Solution

$\begin{array}{l} Let I = \int \frac{\sqrt{\cot x}}{\sin xcos x} dx = \int \frac{\sqrt{\cot x}}{\tan x} \sec^{2} xdx \\ = \int \frac{\sec^{2} x}{(\tan x)^{3 / 2}} dx = \frac{- 2}{\sqrt{\tan x}} + Q = - 2 \sqrt{\cot x} + Q \\ ∴ P = - 2 \Rightarrow P^{4} = 16 \end{array}$

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