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  • Evaluate the definite integrals in Exercises 25 to 33. 30

Evaluate the definite integrals in Exercises 25 to 33.

    Q30.    \int_0^\frac{\pi}{4}\frac{\sin x +\cos x }{9 + 16 \sin 2x}dx

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\int_0^\frac{\pi}{4}\frac{\sin x +\cos x }{9 + 16 \sin 2x}dx

First let's assume t = cosx - sin x so that (sinx +cosx)dx=dt

So,

Now since we are changing the variable, the new limit of the integration will be,

when x = 0, t = cos0-sin0=1-0=1

when x=\pi/4 t=\cos\pi/4-\sin\pi/4=0

Now,

(\cos x-\sin x)^2=t^2

\cos ^2x+\sin^2 x-2\cos x \sin x =t^2

1-\sin 2x =t^2

\sin 2x =1-t^2

Hence our function in terms of t becomes,

\int_{-1}^{0}\frac{dt}{9+16(1-t^2)}=\int_{-1}^{0}\frac{dt}{9+16-16t^2}=\int_{-1}^{0}\frac{dt}{25-16t^2}=\int_{-1}^{0}\frac{dt}{5^2-(4t)^2)}

= \frac{1}{4}\left [\frac{1}{2(5)}\log \frac{5+4t}{5-4t} \right ]_{-1}^0

= \frac{1}{40}\left[ \log (1)-\log (\frac{1}{9})\right ]

=\frac{\log 9}{40}

 

 

Posted by

Pankaj Sanodiya

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