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Good try! Unfortunately, there's a mistake in this piece of integral.😣 Fortunately, it doesn't affect the final answer.:p

No attempt for 2 weeks:( Is it too tedious? I think it's a good practice for algebraic skills.:rolleyes:

Feel free to share your attempt.😜 \int_{0.25}^{0.5}\log_2\left(ex\right)\sin\left(x^x+\ln\sqrt{x}\right)\cos\left(x^x-\ln\sqrt{x}\right)dx...

Has anyone attempted this? \int\frac{x^2-2\pi+\pi^2}{\left(\pi\sin x+x\right)^2+\left(\pi+x\right)\left(\pi-x\right)\cos x+\pi^2\cos^2x}dx =\int\frac{x^2-2\pi+\pi^2}{2\left(x\sin\frac{x}{2}+\pi\cos\frac{x}{2}\right)^2}dx...

After dividing top and bottom by 27^x, the integral is in the form 1/(p^x+q^x+r^x). If this integral has an elementary form, then there must be a relationship between p,q and r. After some trial and error, you would get that relationship.

This integral itself should be quite routine. Getting the final answer in terms of pi is slightly tricky. \int\frac{x\cos x+6x\sin x}{\left(5\sin x-30\cos x+31\right)^2}dx =\frac{1}{5}\int x\frac{\left(5\cos x+30\sin x\right)}{\left(5\sin x-30\cos x+31\right)^2}dx =-\frac{1}{5}\int x\...

Have anyone figured out? The trick is: \int\frac{27^x}{9^x\left(1+\sqrt{5}\right)^x+8^x\left(2+\sqrt{5}\right)^x+6^x\left(3+\sqrt{5}\right)^x}dx...

My method is similar to yours. DI table may make the integration by parts neater. D \qquad I x^2 \qquad \frac{\cos x}{\left(1-\sin x\right)^2} 2x\qquad \frac{1}{1-\sin x} 2\qquad \frac{\cos x}{1-\sin x} 0\qquad -\ln\left(1-\sin x\right) \int_{ }^{ }\frac{x^2\cos x}{\left(1-\sin...

This one should be manageable. \int_0^1\left(\sin^{-1}\frac{x}{x+1}\right)^2dx=\frac{\pi^2}{18}-\frac{\pi}{\sqrt{3}}+2\ln2

This one is challenging if you can't think of a useful substitution. \int\frac{27^x}{9^x\left(1+\sqrt{5}\right)^x+8^x\left(2+\sqrt{5}\right)^x+6^x\left(3+\sqrt{5}\right)^x}dx

This one is tedious. \int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\frac{x^2-2\pi+\pi^2}{\left(\pi\sin x+x\right)^2+\left(\pi+x\right)\left(\pi-x\right)\cos x+\pi^2\cos^2x}dx=\frac{1037\sqrt{2}+259\sqrt{3}-1893}{427}

This one is relatively routine. \int_0^{\frac{\pi}{2}}\frac{x\cos x+6x\sin x}{\left(5\sin x-30\cos x+31\right)^2}dx=\frac{\pi}{72}

This is a rather tough one...especially the final simplification. Feel free to share your attempt...

New integral: \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\log_3\left(\left(1+\pi^x\right)^x\right)}{x^2\sin^2x+\cos^2x+x\sin2x}dx=\frac{4-\pi}{4+\pi}\cdot\frac{\ln\pi}{\ln3}

The answer is: \frac{\ln\left(\sqrt{1+\tan^4x}+\tan^2x\right)}{8}-\frac{\sqrt{1+\tan^4x}}{8\tan^2x}

This one is slightly tedious. \int\left(\csc^22x\right)\left(\csc2x+\cot2x\right)\sqrt{\sin^4x+\cos^4x}dx

It seems no one has attempted yet. This one is a little bit interesting. At the first glance, everything is related to 4^x so it makes sense to substitute u=4^x to get \int_1^4\frac{\left(\sqrt{4-u}\sqrt{7+u^2-5u}+\sqrt[3]{u^2-2u}\right)}{\ln4}du If you put these two functions for 1<u<4 in...

For this integral, some trig identities may make your life easier. Of course, Weierstrass substitution will also work. \int_0^{\pi}\frac{\cos x}{2\sec x-\cos x+2\tan x}dx =\int_0^{\pi}\frac{1}{2\sec^2x-1+2\sec x\tan x}dx =\int_0^{\pi}\frac{1}{\left(\sec x+\tan x\right)^2}dx...

One more...feel free to share your attempt. \int_0^{\pi}\frac{\ln\left(1+\pi^{\cos x}\right)}{2\sec x-\cos x+2\tan x}dx=\left(4-\pi\right)\ln\sqrt{\pi}

Why is it getting so quiet?:cry: Let's have a new integral. Show that \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\ln\left(1+\pi^x\right)}{4\cot x+5\csc x-\sin x}dx=\frac{9-4\sqrt{3}}{36}\ln\pi^{\pi}