Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 73273 by peter frank last updated on 09/Nov/19

Answered by MJS last updated on 09/Nov/19

$$\mathrm{tricky}...$$$$x^2-\cos 2x-1=$$$$=x^2-{2\cos }^{2}x=$$$$=x^2({\sin }^{2}x+{\cos }^{2}x)-{2\cos }^{2}x+\sin x\cos x-\sin x\cos x=$$$$=(x^2{\sin }^{2}x-x\sin x\cos x-{2\cos }^{2}x)+(x\sin x\cos x+x^2{\cos }^{2}x)=$$$$=(x\sin x-2\cos x)(x\sin x+\cos x)+x\cos x(\sin x+x\cos x)$$$$$$$$\Rightarrow$$$$$$$$\int \frac{x^2-\cos 2x-1}{{(x\sin x+\cos x)}^2}dx=$$$$=+\int \frac{x\sin x-2\cos x}{x\sin x+\cos x}dx+\int \frac{x\cos x(\sin x+x\cos x)}{{(x\sin x+\cos x)}^2}dx$$$$$$$$\mathrm{now}\mathrm{the}{2}^{\mathrm{nd}}\mathrm{one}\mathrm{by}\mathrm{parts}$$$$$$$$\int u^{\prime }v=uv-\int {uv}^{\prime }$$$$u^{\prime }=\frac{x\cos x}{{(x\sin x+\cos x)}^2}\to u=-\frac{1}{x\sin x+\cos x}$$$$[t=x\sin x+\cos x\to dx=\frac{dt}{x\cos x}...\int \frac{dt}{t^2}=-\frac{1}{t}]$$$$v=\sin x+x\cos x\to v^{\prime }=-x\sin x+2\cos x$$$$$$$$\int \frac{x\cos x(\sin x+x\cos x)}{{(x\sin x+\cos x)}^2}dx=$$$$=-\frac{\sin x+x\cos x}{x\sin x+\cos x}-\int \frac{x\sin x-2\cos x}{x\sin x+\cos x}dx$$$$$$$$\Rightarrow \mathrm{we}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{have}\mathrm{to}\mathrm{integrate}\mathrm{the}\mathrm{red}\mathrm{one}$$$$$$$$\int \frac{x^2-\cos 2x-1}{{(x\sin x+\cos x)}^2}dx=-\frac{\sin x+x\cos x}{x\sin x+\cos x}+C$$

Commented by mind is power last updated on 10/Nov/19

$$\mathrm{nice}\mathrm{Sir}$$

Commented by peter frank last updated on 10/Nov/19

$$thankyou$$

Commented by peter frank last updated on 10/Nov/19

$$sorrysirwhywe{don}^{{}^{\prime }}tintegrateredone$$

Commented by MJS last updated on 10/Nov/19

$$\int \frac{x^2-\cos 2x-1}{{(x\sin x+\cos x)}^2}dx=$$$$=+\int \frac{x\sin x-2\cos x}{x\sin x+\cos x}dx+\int \frac{x\cos x(\sin x+x\cos x)}{{(x\sin x+\cos x)}^2}dx$$$$\mathrm{with}\int \frac{x\cos x(\sin x+x\cos x)}{{(x\sin x+\cos x)}^2}dx=$$$$=-\frac{\sin x+x\cos x}{x\sin x+\cos x}-\int \frac{x\sin x-2\cos x}{x\sin x+\cos x}dx$$$$\mathrm{gives}$$$$+\int \frac{x\sin x-2\cos x}{x\sin x+\cos x}dx-\frac{\sin x+x\cos x}{x\sin x+\cos x}-\int \frac{x\sin x-2\cos x}{x\sin x+\cos x}dx=$$$$=-\frac{\sin x+x\cos x}{x\sin x+\cos x}$$

Commented by peter frank last updated on 10/Nov/19

$$thankyou$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com