If $A_n=\int_{\frac{\pi}{2}} e^{-x} \cdot \cos ^n x d x$, then $\frac{A_4-A_6}{A_4}=$

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Aicharya · Accepted Answer

We have, \(\begin{aligned} & A_6=\int_{\frac{\pi}{2}}^{\infty} e^{-x} \cos ^6 x d x \ & =\left[-e^{-x} \cos ^6 x\right]_{\frac{\pi}{2}}^{\infty}-\int_{\frac{\pi}{2}}^{\infty}\left(-e^{-x}\right) 6 \cos ^5 x(-\sin x) d x \ & =0-6 \int_{\frac{\pi}{2}}^{\infty} e^{-x} \cos ^5 x \sin x d x \ & \Rightarrow \quad \frac{A_6}{-6}=\int_{\frac{\pi}{2}}^{\infty} e^{-x} \cos ^5 x \sin x d x \ & \Rightarrow \quad \frac{-1}{6} A_6=\left[-e^{-x} \cos ^5 x \sin x\right]_{\frac{\pi}{2}}^{\infty} \ & -\int_{\frac{\pi}{2}}^{\infty}\left(-e^{-x}\right)\left[5 \cos ^4 x(-\sin x) \sin x+\cos ^5 x \cdot \cos x\right] d x \ & \Rightarrow \frac{-1}{6} A_6=0+\int_{\frac{\pi}{2}}^{\infty} e^{-x}\left[-5 \cos ^4 x\left(1-\cos ^2 x\right)+\cos ^6 x\right] d x \ & \Rightarrow \frac{-1}{6} A_6=-5 \int_{\frac{\pi}{2}}^{\infty} e^{-x} \cos ^4 x d x+6 \int_{\frac{\pi}{2}}^{\infty} e^{-x} \cos ^6 x d x \ & \Rightarrow \quad \frac{-1}{6} A_6=-5 A_4+6 A_6 \ & \Rightarrow \quad 5 A_4=\frac{37}{6} A_6 \ & \Rightarrow \quad A_6=\frac{30}{37} A_4 \ & \therefore \quad \frac{A_4-A_6}{A_4}=\frac{A_4-\frac{30}{37} A_4}{A_4}=\frac{7}{37} \end{aligned}\)

If \(A_n=\int_{\frac{\pi}{2}} e^{-x} \cdot \cos ^n x d x\), then \(\frac{A_4-A_6}{A_4}=\)

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