3
$\begingroup$

What results?

$$\int^{\frac{\pi}{2}}_0 \frac{\sin x}{\sin x+\cos x+1}dx$$

my try : $u= \tan \frac{x}{2 } $

but : What is the short way?

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    Substitution $x=\dfrac{\pi}{2}-u$. – 2017-01-21
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    i think the $$\tan(\frac{x}{2})$$ is the right way – 2017-01-21
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    a possible result for the indefinite integral is $$\frac{x}{2}-\log \left(\sin \left(\frac{x}{2}\right)+\cos \left(\frac{x}{2}\right)\right)$$ – 2017-01-21

2 Answers 2

3

\begin{eqnarray} \sin x+\cos x+1 &=& 1+\sin x+\cos x \\ &=& (\sin\frac{x}{2}+\cos\frac{x}{2})^2+(\cos^2\frac{x}{2}-\sin^2\frac{x}{2}) \\ &=& (\sin\frac{x}{2}+\cos\frac{x}{2})2\cos\frac{x}{2} \end{eqnarray} We have by substitution $x=\dfrac{\pi}{2}-u$ \begin{eqnarray} \int^{\frac{\pi}{2}}_0 \frac{\sin u}{\sin u+\cos u+1}du &=& \int^{\frac{\pi}{2}}_0 \frac{\cos x}{\sin x+\cos x+1}dx\\ &=& \int^{\frac{\pi}{2}}_0 \frac{\cos^2\frac{x}{2}-\sin^2\frac{x}{2}}{(\sin\frac{x}{2}+\cos\frac{x}{2})2\cos\frac{x}{2}}dx\\ &=& \frac12\int^{\frac{\pi}{2}}_0 \frac{(\cos\frac{x}{2}-\sin\frac{x}{2})}{\cos\frac{x}{2}}dx\\ &=& \frac12\int^{\frac{\pi}{2}}_0 1-\tan\frac{x}{2}dx\\ &=& \frac{\pi}{4}+\ln\cos\frac{x}{2}\Big|_0^\frac{\pi}{2}\\ &=& \color{blue}{\frac{\pi}{4}-\frac12\ln2} \end{eqnarray}

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    @DavidQuinn Final number is corrected. Thank you for your attention. – 2017-01-22
4

Hint... You can write the integral as $$I=\int_0^\frac{\pi}{2} \frac{\sin\frac x2}{\sin\frac x2+\cos \frac x2} dx$$

Then consider $$J=\int_0^\frac{\pi}{2} \frac{\cos\frac x2}{\sin\frac x2+\cos \frac x2} dx$$

Now work out $J-I$ and $J+I$

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    Unfortunately, with substitution upper limit in $J$ will change to $\dfrac{\pi}{4}$. – 2017-01-21
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    @MyGlasses...What substitution? – 2017-01-21
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    $I+J=\int_0^\frac{\pi}{2} \frac{\cos\frac x2+\sin \frac{x}{2}}{\sin\frac x2+\cos \frac x2} dx=\int_0^\frac{\pi}{2} dx$ $J-I=\int_0^\frac{\pi}{2} \frac{\cos\frac x2-\sin \frac{x}{2}}{\sin\frac x2+\cos \frac x2} dx$ now??!! – 2017-01-21
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    The first primitive is $x$ and the second is $2\ln|\sin\frac x2+\cos\frac x2|$ – 2017-01-21
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    @MyGlasses. I have used half-angle formulas to change the original integrand. – 2017-01-22