If ∫$$cos3$$⁡$$xsin11$$⁡$$xdx=$$−$$2Atan$$−$$9/2$$⁡$$x+Btan$$−$$5/2$$⁡$$x+C$$ then $$1A+1B$$ is equal to

Question

If ∫$$cos3$$⁡$$xsin11$$⁡$$xdx=$$−$$2Atan$$−$$9/2$$⁡$$x+Btan$$−$$5/2$$⁡$$x+C$$  then $$1A+1B$$ is equal to

Infinity Learn · Accepted Answer

We observe that the sum of the exponents of cos x and $sinx i s (\frac{3}{2} - \frac{11}{2}) = - 4 = a$ ,negative even integer.

So, we divide numerator and denominator by cos⁴ x

$\begin{array}{l} ∴ I = \int \sqrt{\frac{\cos^{3} x}{\sin^{11} x}} dx = \int \frac{\cos^{3 / 2} x}{\sin^{11 / 2} x} dx \\ \Rightarrow I = \int \frac{\cos^{3 / 2} x \times \cos^{4} x}{\sin^{11 / 2} x} \times \frac{1}{\cos^{4} x} dx \\ \Rightarrow I = \int \frac{1}{\tan^{11 / 2} x} (1 + \tan^{2} x) d (\tan x) \\ \Rightarrow I = \int (\tan^{- 11 / 2} x + \tan^{- 7 / 2} x) d (\tan x) \\ \Rightarrow I = - \frac{2}{9} \tan^{- 9 / 2} x - \frac{2}{5} \tan^{- 5 / 2} x + C \\ \Rightarrow I = - 2 \{\frac{1}{9} \tan^{- 9 / 2} x + \frac{1}{5} \tan^{- 5 / 2} x\} + C \end{array}$

Hence , $A = \frac{1}{9} and B = \frac{1}{5}$

$∴ \frac{1}{A} + \frac{1}{B} = 14$

Methods of integration

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If $\int \sqrt{\frac{\cos^{3} x}{\sin^{11} x}} dx = - 2 \{{Atan}^{- 9 / 2} x + {Btan}^{- 5 / 2} x\} + C$ then $\frac{1}{A} + \frac{1}{B}$ is equal to

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Methods of integration

Remember concepts with our Masterclasses.

If ∫cos3⁡xsin11⁡xdx=−2Atan−9/2⁡x+Btan−5/2⁡x+C then 1A+1B is equal to

Ready to Test Your Skills?

free study material

If $\int \sqrt{\frac{\cos^{3} x}{\sin^{11} x}} dx = - 2 \{{Atan}^{- 9 / 2} x + {Btan}^{- 5 / 2} x\} + C$ then $\frac{1}{A} + \frac{1}{B}$ is equal to