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Calculate:

$$\int _{0}^{\pi }\cos(x)\log(\sin^2 (x)+1)dx$$

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closed as off-topic by heropup, Claude Leibovici, probablyme, user26857, G. Sassatelli Feb 6 at 9:10

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3  
Make the substitution $u = \pi - x$ and see what follows. No complicated calculations necessary. – Daniel Fischer Feb 5 at 16:47

Behold the power of symmetry:

$$\cos(\pi - x) = - \cos x\quad\text{and} \quad \sin (\pi - x) = \sin x,$$

therefore

\begin{align} \int_0^\pi \cos x \log (\sin^2 x + 1) \,dx &= \int_0^\pi \cos (\pi - u)\log (\sin^2(\pi - u) + 1)\,du\\ &= - \int_0^\pi \cos u\log (\sin^2 u + 1)\,du, \end{align}

hence the integral evaluates to $0$.

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nice..............+1 – Bhaskara-III Feb 5 at 18:12
2  
You were inspired by Daniel Fischer's comment. ;-)) +1 – Dr. MV Feb 5 at 19:55

$$\int\cos x\ln(1+\sin^2x)\ dx$$

$$=\ln(1+\sin^2x)\int\cos x\ dx-\int\left(\dfrac{d\ \ln(1+\sin^2x)}{dx}\int\cos x\ dx\right)dx$$

$$=\sin x\cdot\ln(1+\sin^2x)-\int\dfrac{2\sin^2x\cos x}{1+\sin^2x}dx$$

$$\int\dfrac{2\sin^2x\cos x}{1+\sin^2x}dx=\int\dfrac{2(1+\sin^2x-1)\cos x}{1+\sin^2x}dx=2\int\cos x\ dx-2\int\dfrac{\cos x}{1+\sin^2x}dx$$

Set $\sin x=u$ for the last integral

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$$\int_0^\pi \cos(x) \log(\sin^2(x)+1)dx$$

The first thing we want to do is to remove the $\log$ from the integral. Performing integration by parts we set $u=\log(\sin^2(x)+1)$ and $dv=\cos(x)dx$ to find $du = \frac{2\sin(x)\cos(x)}{\sin^2(x)+1} dx$ and $v = \sin(x)$.

Thus the integral transforms to $$\left.\sin(x)\log(\sin^2(x)+1)\right|_0^\pi - \int_0^\pi \frac{2\sin^2(x)\cos(x)}{\sin^2(x)+1} dx=0- \int_0^\pi \frac{2\sin^2(x)\cos(x)}{\sin^2(x)+1} dx$$

Now let us shift the integral by $\pi/2$, replace $x=u+\pi/2$ to find:

$$-\int_0^\pi \frac{2\sin^2(x)\cos(x)}{\sin^2(x)+1} dx = -\int_{-\pi/2}^{\pi/2} \frac{2\sin^2(u+\pi/2)\cos(u+\pi/2)}{\sin^2(u+\pi/2)+1} dx = \int_{-\pi/2}^{\pi/2} \frac{2\cos^2(u)\sin(u)}{\cos^2(u)+1} dx$$

Now notice that $\frac{2\cos^2(x)}{\cos^2(x)+1}$ is even and $\sin(x)$ is odd, thus their product is odd. The integration is happening over a symmetric interval, so $$\int_{-\pi/2}^{\pi/2} \frac{2\cos^2(u)\sin(u)}{\cos^2(u)+1} dx=0.$$

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This answer is in the same vein as Daniel Fischer's, but presented differently.

Make the substitution $u = x - \pi/2$. Then the integral becomes $$-\int_{-\pi/2}^{\pi/2} \sin u \log(1 + \cos^2 u) \, du,$$ which is the integral of an odd function over $[-\pi/2,\pi/2]$. Therefore the integral is zero.

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HINT: Substitute $y=\sin(x), y' = \cos(x)$. Then use $w^2+1 = (1+iw)(1-iw)$ and logarithm laws. The integral over logarithm is easy to find in integral tables.

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