MHT-CET Full Test-7 Mathematics Que-49 Solution

Q.49. If \int \cos^3{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) equals to

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) .

Explanation :-

Let I = \int \cos^{\frac{3}{5}}{x} \sin^3{x} \, dx .

I = \int \cos^{\frac{3}{5}}{x} (1 - \cos^2{x}) \sin{x} \, dx .

I = \int \cos^{\frac{3}{5}}{x} \sin{x} \, dx - \int \cos^{\frac{13}{5}}{x} \sin{x} \, dx .

Let \cos{x} = t \Rightarrow -\sin{x} \, dx = dt .

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

Comparing with \frac{-1}{m} \cos^m x + \frac{1}{n} \cos^n x + c , we get m = \frac{8}{5}, n = \frac{18}{5}

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