I want to find $$\int \frac{x+x\sin x+e^x \cos x}{e^x+x\cos x-e^{x} \sin x} dx.$$ But since algebraic, exponential and trigonometric functions are involved I am not able to solve it. Please help in finding it by hand.
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Wolframalpha's answer – Arctic Char Jul 24 '20 at 10:40
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But how to get it by hand. – Dharmendra Singh Jul 24 '20 at 10:44
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Wow, nice @ArcticChar. How might one notice / spot the substitution? Seems hard... – Benjamin Wang Jul 24 '20 at 10:46
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Dharmendra Singh, nice profile! Would you mind joining us here https://chat.meta.stackexchange.com/rooms/1508/general-chatroom – Arjun Jul 31 '20 at 12:40
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Note the law of equal peopoetions, we have: $$\frac{1+\sin x}{\cos x}=\frac{\cos x}{1-\sin x} \implies \frac{x(1+\sin x)}{x \cos x}=\frac{e^x \cos x}{e^x-e^x \sin x}=\frac{x+x\sin x+e^x \cos x}{x\cos x+ e^{x}-e^{x} \sin x}$$ So $$\int \frac{x+x\sin x+e^x \cos x}{e^x+x\cos x-e^{x} \sin x} dx= \int \frac{\cos x}{1-\sin x}dx=-\ln (1-\sin x)+C$$
Z Ahmed
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+1 I believe this is more commonly known as componendo and dividendo. – Toby Mak Jul 24 '20 at 11:09
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It's because if $\frac{a}{b}=\frac{c}{d}$ (and $b+d\ne0$), then $\frac{a}{b}=\frac{a+c}{b+d}$. – Minus One-Twelfth Jul 24 '20 at 11:15
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Hint:
For $1+\sin x\ne0,$
the denominator
$$=e^x(1-\sin x)+x\cos x$$
$$=\dfrac{e^x\cos^2x+x\cos x(1+\sin x)}{1+\sin x}$$
$$=\dfrac{\cos x}{1+\sin x}\cdot (e^x\cos x+x(1+\sin x))$$
Can you recognize the numerator as a multiplicand?
lab bhattacharjee
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