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Question Number 193526 by SaRahAli last updated on 16/Jun/23

Answered by MM42 last updated on 16/Jun/23

$$I=\int \frac{sinx\times cosx}{sinx+{cos}^{2}x}dx=-\int \frac{sinx\times cosx}{sinx-sinx-1}dx$$$$letsinx=u\Rightarrow cosxdx=du$$$$\Rightarrow I=-\int \frac{udu}{u^2-u-1}du=\int \frac{A}{u-u_1}du+\int \frac{B}{u-u_2}du$$$$thesolutionisverysimple$$

Answered by Subhi last updated on 16/Jun/23

$$\int \frac{sin\left(x\right)}{tan\left(x\right)+cos\left(x\right)}dx\Rrightarrow \int \frac{sin\left(x\right)}{\frac{sin\left(x\right)}{cos\left(x\right)}+cos\left(x\right)}$$$$\int \frac{sin\left(x\right).cos\left(x\right)}{sin\left(x\right)+1-{sin}^2\left(x\right)}\Rrightarrow {sin}^2\left(x\right)=u\Rrightarrow 2sin\left(x\right)cos\left(x\right)dx=du$$$$\frac{-1}{2}\int \frac{du}{u-\sqrt{u}-1}\Rrightarrow \sqrt{u}=v\Rrightarrow \frac{1}{2\sqrt{u}}du=dv$$$$\frac{-1}{2}\int \frac{2v}{v^2-v-1}dv\Rrightarrow -\int \frac{v}{v^2-v-1}\Rrightarrow -\int \frac{a}{v-\frac{1+\sqrt{5}}{2}}+\frac{b}{v-\frac{1-\sqrt{5}}{2}}$$$$av-a.\frac{1-\sqrt{5}}{2}+bv-b.\frac{1+\sqrt{5}}{2}$$$$a+b=1\looparrowright -\frac{1-\sqrt{5}}{2}a-\frac{1+\sqrt{5}}{2}b=0$$$$\sqrt{5}a=\frac{1+\sqrt{5}}{2}\Rrightarrow a=\frac{5+\sqrt{5}}{10}\Rrightarrow b=\frac{5-\sqrt{5}}{10}$$$$-\frac{5+\sqrt{5}}{10}\int \frac{1}{v-\frac{1+\sqrt{5}}{2}}-\frac{5-\sqrt{5}}{10}\int \frac{1}{v-\frac{1-\sqrt{5}}{2}}$$$$-\frac{5+\sqrt{5}}{10}.ln\mid sin\left(x\right)-\frac{1+\sqrt{5}}{2}\mid -\frac{5-\sqrt{5}}{10}.ln\mid sin\left(x\right)-\frac{1-\sqrt{5}}{2}\mid$$

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