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Question Number 200606 by Calculusboy last updated on 20/Nov/23

Answered by Frix last updated on 20/Nov/23

$$\int \frac{\cos x{\sin }^{2}x}{\cos x+\sin x}dx\stackrel{t=x-\frac{\pi }{4}}{=}$$$$=\int (\frac{1}{4}+\frac{\cos t\sin t}{2}-\frac{{\sin }^{2}t}{2}-\frac{\tan t}{4})dt=$$$$=\frac{t}{4}-\frac{{\cos }^{2}t}{4}-\frac{t+\cos t\sin t}{4}+\frac{\ln \cos t}{4}=$$$$=\frac{\ln \mid \cos x+\sin x\mid }{4}-\frac{(\cos x+\sin x)\cos x}{4}+C$$

Commented by Calculusboy last updated on 21/Nov/23

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{s}\mathit{s}\mathit{i}\mathit{r}$$

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