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Question Number 144059 by 0731619 last updated on 21/Jun/21

Answered by Olaf_Thorendsen last updated on 21/Jun/21

$$\mathrm{F}\left(x\right)=\int \frac{{\sin }^{2}x}{b^2{\cos }^{2}x+a^2}dx$$$$\mathrm{F}\left(x\right)=\int \frac{{\tan }^{2}x}{b^2+\frac{a^2}{{\cos }^{2}x}}dx$$$$\mathrm{F}\left(x\right)=\int \frac{{\tan }^{2}x}{b^2+a^2(1+{\tan }^{2}x)}dx$$$$\mathrm{Let}t=\tan x$$$$\mathrm{F}\left(t\right)=\int \frac{t^2}{b^2+a^2(1+t^2)}.\frac{dt}{1+t^2}$$$$\mathrm{F}\left(t\right)=\int (\frac{\frac{a^2+b^2}{b^2}}{a^2{t}^2+a^2+b^2}-\frac{\frac{1}{b^2}}{1+t^2})dt$$$$\mathrm{F}\left(t\right)=\frac{a^2+b^2}{a^2{b}^2}\int \frac{dt}{t^2+\frac{a^2+b^2}{a^2}}-\frac{1}{b^2}\int \frac{dt}{1+t^2}$$$$\mathrm{F}\left(t\right)=\frac{a^2+b^2}{a^2{b}^2}\left(\frac{1}{\sqrt{\frac{a^2+b^2}{a^2}}}\arctan \frac{t}{\sqrt{\frac{a^2+b^2}{a^2}}}\right)$$$$-\frac{1}{b^2}\arctan t+\mathrm{C}$$$$\mathrm{F}\left(t\right)=\frac{\sqrt{a^2+b^2}}{{ab}^2}\arctan \left(\frac{at}{\sqrt{a^2+b^2}}\right)-\frac{1}{b^2}\arctan t+\mathrm{C}$$$$\mathrm{F}\left(x\right)=\frac{\sqrt{a^2+b^2}}{{ab}^2}\arctan \left(\frac{a.\tan x}{\sqrt{a^2+b^2}}\right)-\frac{x}{b^2}+\mathrm{C}$$

Commented by 0731619 last updated on 21/Jun/21

$$thanks$$

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