Question

Sinx Cosx
Evaluate :
dx
16+ 9 sin 2x
0​

Answer 1

Sin x cos = 1hgafebJjajBsgwghsb

Answer 2

Given,

$\sf\displaystyle\int \dfrac{sinx.cosx}{9+16sin2x} \ dx\\\\=\int \dfrac{sinx.cosx dx}{9-16((sinx-cosx)^2-1)}\\\\=\int \dfrac{sinx.cosxdx}{25-16(sinx-cosx)^2}$

Let sinx - cosx = t

$= \sf \displaystyle \int \dfrac{dt}{25-(4t)^2}\\\\=\dfrac{\dfrac{1}{10}ln\left(\dfrac{4t+5}{4t-5}\right)}{4} + C\\\\=\dfrac{1}{40} ln\left(\dfrac{4(sinx-cosx)+5}{4(sinx+cosx)-5}\right) + C$