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Question Number 216706 by issac last updated on 16/Feb/25

$$\int \frac{\mathrm{d}z}{1+\sin \left(z\right)\cos \left(z\right)}=$$$$\int \frac{{\sec }^2\left(z\right)\mathrm{d}z}{{\sec }^2\left(z\right)+\tan \left(z\right)}\mathrm{multiply}{\sec }^2\left(z\right)$$$${\sec }^2\left(z\right)=1+{\tan }^2\left(z\right)$$$$\int \frac{{\sec }^2\left(z\right)\mathrm{d}z}{1+\tan \left(z\right)+{\tan }^2\left(z\right)}$$$$s=\tan \left(z\right)$$$$\mathrm{d}s={\sec }^2\left(z\right)\mathrm{d}z$$$$\int \frac{\mathrm{d}s}{s^2+s+1}=\int \frac{\mathrm{d}s}{{(s+\frac{1}{2})}^2+\frac{3}{4}}=\int \frac{\mathrm{d}q}{q^2+\frac{3}{4}}$$$$q=s+\frac{1}{2}$$$$\frac{2}{\sqrt{3}}\int \frac{1}{w^2+1}\mathrm{d}w$$$$w=\frac{2q}{\sqrt{3}}$$$$\frac{2}{\sqrt{3}}{\tan }^{-1}\left(w\right)+C\to \frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2q}{\sqrt{3}}\right)+C$$$$q=s+\frac{1}{4}$$$$\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2}{\sqrt{3}}(s+\frac{1}{2})\right)+C$$$$s=\tan \left(z\right)$$$$\therefore \int \frac{\mathrm{d}z}{1+\sin \left(z\right)\cos \left(z\right)}=\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2\tan \left(z\right)+1}{\sqrt{3}}\right)+C$$$${\int }_0^{2\pi }=\frac{4\pi }{\sqrt{3}}$$

Commented by Ghisom last updated on 16/Feb/25

why can't you just use this app as intended, without giving "holy shits" and opening new questions for answers of old questions? you're embarrassing yourself, that's all. who and what do you think you are?

Commented by issac last updated on 16/Feb/25

$$????$$

Commented by Frix last updated on 16/Feb/25

$$!!!!$$

Commented by MathematicalUser2357 last updated on 29/Mar/25

$$\mathrm{Correct}...!(\mathrm{Maybe}\mathrm{you}\mathrm{hid}\mathrm{the}\mathrm{integrand}\mathrm{because}\mathrm{of}\mathrm{time}\mathrm{comsumption}.)$$$$\mathrm{Feedback}:$$$$\mathrm{Great}\mathrm{job}!\mathrm{You}\mathrm{learned}\mathrm{derivatives},\mathrm{right}?$$$$\mathrm{But},\mathrm{I}\mathrm{think}\mathrm{you}\mathrm{forgot}\mathrm{the}\mathrm{notation}\mathrm{somedays}.$$$$\mathrm{The}\mathrm{derivative}\mathrm{notations}\mathrm{are}:\frac{dy}{dx},\frac{d}{dx}\left(y\right),y^{\prime }.$$$$\mathrm{Hope}\mathrm{you}\mathrm{like}\mathrm{this}\mathrm{feedback}!$$

Answered by issac last updated on 16/Feb/25

$$\mathrm{q}216695$$

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