Test Index

AP EAMCET Engineering 21 Apr 2019 Shift 2 Paper

Section: Mathematics

© examsnet.com

Question : 77 of 160

Marks: +1, -0

\int\limits_{0}^{\infty} e^{-x} \sin^6 x

dx =

$\frac{24}{85}$
$\frac{124}{285}$
$\frac{136}{529}$
$\frac{144}{629}$

Solution:

The formula for the integration,

I_{m}=\int\limits_{0}^{\infty} e^{-x} \sin ^{m} x \, dx,

where

m>2

then

I_{m}=\;\frac{m(m-1)}{1+m^{2}} I_{m-2}

\text{Now}

I_{6}=\int\limits_{0}^{\infty} e^{-x} \sin 6 x \, dx

I_{6}=\;\frac{6(5)}{1+6^{2}} I_{4}

I_{6}=\;\frac{30}{37} I_{4}

Similarly,

I_{4}=\int\limits_{0}^{\infty} e^{-x} \sin 4 x \, dx

I_{4}=\;\frac{4(3)}{1+4^{2}} I_{2}

I_{4}=\;\frac{12}{17} I_{2}

Similarly

I_{6}=\int\limits_{0}^{\infty} e^{-x} \sin 6 x \, dx

I_{2}=\;\frac{2(1)}{1+2^{2}} I_{0}

I_{2}=\;\frac{2}{5} I_{0}

And

I_{0}=\int\limits_{0}^{\infty} e^{-x} \sin ^{0} x \, dx

I_{0}=[-e^{-x}]_{0}^{\infty}=1

The value of

I_{6}

is,

I_{6}=\;\frac{30}{37} \times \;\frac{12}{17} \times \;\frac{2}{5} \times 1=\;\frac{144}{629}

© examsnet.com

Go to Question:

Prev Question

Next Question