sinxcosx-4-sin-4-x-dx-

Question Number 9520 by ridwan balatif last updated on 12/Dec/16

\int \frac{sinxcosx}{4 + \sin^{4} x} dx = \dots ?

Answered by mrW last updated on 12/Dec/16

u = \sin x

du = \cos xdx

\int \frac{sinxcosx}{4 + \sin^{4} x} dx = \int \frac{u}{4 + u^{4}} du

v = u^{2}

dv = 2 udu

\int \frac{u}{4 + u^{4}} du = \frac{1}{2} \int \frac{dv}{2^{2} + v^{2}} = \frac{1}{2} \times \frac{1}{2} \tan^{- 1} (\frac{v}{2}) + C

= \frac{1}{4} \tan^{- 1} (\frac{u^{2}}{2}) + C

= \frac{1}{4} \tan^{- 1} (\frac{\sin^{2} x}{2}) + C

Commented by ridwan balatif last updated on 12/Dec/16

thank you sir