Question

Integration by Partial Fractions

∫dx/(3-5Sin2x) from 0 to π/4 

Answer 1

Given integral is-

$\int_0^{\frac{\pi}{4}} \dfrac{dx}{3-5sin2x}$

Let u=2x, du=2dx

$=\dfrac{1}{2}\int_0^{\frac{\pi}{2}} \dfrac{1}{3-5sinu}du$

Let $v=tan(\dfrac{u}{2})$

      

SO Above integral becomes-

$=\dfrac{1}{2}\int_0^1 \dfrac{2}{3(1+v^2)-10v}$

$=\dfrac{1}{2}.2.\int_0^1\dfrac{1}{3(1+v^2)-10v}dv$

Taking the partial fraction-

$=\dfrac{1}{2}.2\int_0^1[ -\dfrac{3}{8(3v-1)}+\dfrac{1}{8(v-3)}]dv$

$=\dfrac{1}{2}.2.(-\dfrac{1}{8}ln|3v-1|+\dfrac{1}{8}ln|v-3|)|_0^1$

$=\dfrac{1}{2}.2(\dfrac{1}{8}.ln\dfrac{v-3}{3v-1})|_0^1$

$=\dfrac{1}{8}(ln\dfrac{-2}{2}-ln3)$

$=\dfrac{ln(-1)-ln3}{8}$