Question

Evaluate the definite integral: $\int_{0}^{\frac{\pi}{4}}\frac{sinx+cosx}{9+16sin2x}dx$

Answer 1

$Let\space I=\int_{0}^{\frac{\pi}{4}}\frac{sinx+cosx}{9+16sin2x}dx$

Also,let $sinx-cosx=t⇒(cosx+sinx)dx=dt$

When$x=0,t=-1$ and when $x=\frac{\pi}{4},t=0$

$⇒(sinx-cosx)^{2}=t^{2}$

$⇒sin^{2}x+cos^{2}x-2sinxcosx=t^{2}$

$⇒1-sin2x=t^{2}$

$⇒sin2x=1-t^{2}$

$∴I=\int_{-1}^{0}\frac{dt}{9+16(1-t^{2})}$

$=\int_{-1}^{0}\frac{dt}{9+16-16t^{2}}$

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

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

$=\frac{1}{40}[log(1)-log|\frac{1}{9}|]$

$=\frac{1}{40}log9$