MHT-CET Full Test-8 Mathematics Que-49 Solution

Q.49 \int \cos^{\frac{3}{5}} x \cdot \sin^3 x \, dx = \frac{-1}{m} \cos^m x + \frac{1}{n} \cos^n x + c

(where c is the constant of integration, then (m, n) =

A. \left(\frac{18}{5}, \frac{8}{5}\right)

B. \left(\frac{-8}{5}, \frac{18}{5}\right)

C. \left(\frac{8}{5}, \frac{18}{5}\right)

D. \left(\frac{-18}{5}, \frac{-8}{5}\right)

Answer: C. \left(\frac{8}{5}, \frac{18}{5}\right)

\text{Let } I = \int \cos^{\frac{3}{5}} x \sin^3 x \, dx = \int \cos^{\frac{3}{5}} x \left(1 - \cos^2 x\right) \sin x \, dx = \int \cos^{\frac{3}{5}} x \sin x \, dx - \int \cos^{\frac{13}{5}} x \sin x \, dx \text{Let } \cos x = t \Rightarrow -\sin x \, dx = dt \therefore I = - \int t^{\frac{3}{5}} dt + \int t^{\frac{13}{5}} dt = \frac{-1}{\left(\frac{8}{5}\right)} t^{\frac{8}{5}} + \frac{1}{\left(\frac{18}{5}\right)} t^{\frac{13}{5}} + c = \frac{-1}{\left(\frac{8}{5}\right)} \cos^{\frac{8}{5}} x + \frac{1}{\left(\frac{18}{5}\right)} \cos^{\frac{13}{5}} x + c \text{Comparing with } \frac{-1}{m} \cos^m x + \frac{1}{n} \cos^n x + c, \text{ we get } m = \frac{8}{5}, n = \frac{18}{5}
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