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$$ \int_0^{\pi/2} \frac{\sin\theta}{1+\cos\theta} \, d\theta $$

My working thus far:

$$u=1+\cos\theta$$

$$\text{d}u=-\sin\theta \ \text{d} \theta$$

Substituting limits in and obtaining them in terms of u:

$$\int^1_2 \frac{\sin\theta}{u} \cdot \frac{-1}{\sin\theta} \ \text{d}u$$

The answer in the back is $\ln(3/2)$. From my limits, I haven't done this right. Where am I going wrong?

Michael Hardy
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Jim
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4 Answers4

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I think book has printing mistake. You are doing well and all steps are correct

DSinghvi
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$$ \int_2^1 \frac{\sin\theta}{u}\cdot \frac{-1}{\sin\theta}\, d\theta = \int_2^1 \frac{-1}{u}\,du = \int_1^2 \frac 1 u \, du = \cdots\cdots $$

($\ln(3/2)$ is incorrect.)

Michael Hardy
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No need, really, to do substitution. Directly, and using that

$$\int\frac{f'(x)}{f(x)}dx=\log |f(x)|+C$$

we get

$$\int\frac{\sin x}{1+\cos x}dx=-\int\frac{(1+\cos x)'}{1+\cos x}dx=-\log|1+\cos x|+C\;\ldots$$

Timbuc
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  • I wonder if this should be considered different from substitution? $\displaystyle\int\underbrace{{}\ \frac{1}{f(x)}\ {}}{1/u} \Big( ,\underbrace{f'(x) , dx}{du} ,\Big)$, etc. How would you know that $\displaystyle\int \frac{1}{f(x)},f'(x),dx = \log|f(x)|+C$ without substitution, i.e. without the chain rule? – Michael Hardy Jun 05 '14 at 20:33
  • You have a point there @MichaelHardy, yet using this method you don't have to change limits when dealing with definite integrals. For me at least that's a huge advantage. – Timbuc Jun 05 '14 at 22:37
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If $f(x) = -\ln (1 + \cos x)$, then $f'(x) = {\sin x \over 1 + \cos x}$.

Consequently, $\int_0^{\pi \over 2} f'(x) dx = f({\pi \over 2})-f(0) = \ln 2$.

copper.hat
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