Q99 / 150 · MHT CET 2023 (12 May Shift 1) · Mathematics
If $\int \cos ^{\frac{3}{5}} x \cdot \sin ^3 x \mathrm{~d} x=\frac{-1}{\mathrm{~m}} \cos ^{\mathrm{m}} x+\frac{1}{\mathrm{n}} \cos ^{\mathrm{n}} x+\mathrm{c}$, (where $c$ is the constant of integration), then $(\mathrm{m}, \mathrm{n})=$
- A$\left(\frac{18}{5}, \frac{8}{5}\right)$
- B$\left(\frac{-8}{5}, \frac{18}{5}\right)$
- Correct answerC$\left(\frac{8}{5}, \frac{18}{5}\right)$
- D$\left(\frac{-18}{5}, \frac{-8}{5}\right)$
Solution
Let
$$
\begin{aligned}
\mathbf{I} & =\int \cos ^{\frac{3}{5}} x \sin ^3 x \mathrm{~d} x \\
& =\int \cos ^{\frac{3}{5}} x\left(1-\cos ^2 x\right) \sin x \mathrm{~d} x \\
& =\int \cos ^{\frac{3}{5}} x \sin x \mathrm{~d} x-\int \cos ^{\frac{13}{5}} x \sin x \mathrm{~d} x
\end{aligned}
$$
Let $\cos x=\mathrm{t} \Rightarrow-\sin x \mathrm{~d} x=\mathrm{dt}$
$$
\begin{aligned}
\therefore \quad I & =-\int t^{\frac{3}{5}} d t+\int t^{\frac{13}{5}} d t \\
& =\frac{-1}{\left(\frac{8}{5}\right)} t^{\frac{8}{5}}+\frac{1}{\left(\frac{18}{5}\right)} t^{\frac{13}{5}}+c \\
& =\frac{-1}{\left(\frac{8}{5}\right)} \cos ^{\frac{8}{5}} x+\frac{1}{\left(\frac{18}{5}\right)} \cos ^{\frac{13}{5}} x+\mathrm{c}
\end{aligned}
$$
Comparing with $\frac{-1}{\mathrm{~m}} \cos ^{\mathrm{m}} x+\frac{1}{\mathrm{n}} \cos ^{\mathrm{n}} x+\mathrm{c}$, we get $\mathrm{m}=\frac{8}{5}, \mathrm{n}=\frac{18}{5}$
Hence, the correct answer is option: 'C'.